Square root, Representation and Logarithm of complex numbers Questions in English

(C) Let $\sqrt{3 + \sqrt{5}} = \sqrt{x} + \sqrt{y}$.
Squaring both sides,we get $3 + \sqrt{5} = x + y + 2\sqrt{xy}$.
Comparing the rational and irrational parts,we have $x + y = 3$ and $2\sqrt{xy} = \sqrt{5}$,which implies $4xy = 5$,so $xy = \frac{5}{4}$.
We know that $(x - y)^2 = (x + y)^2 - 4xy = 3^2 - 4(\frac{5}{4}) = 9 - 5 = 4$.
Thus,$x - y = 2$.
Solving the system $x + y = 3$ and $x - y = 2$,we get $2x = 5 \implies x = \frac{5}{2}$ and $2y = 1 \implies y = \frac{1}{2}$.
Therefore,$\sqrt{3 + \sqrt{5}} = \sqrt{\frac{5}{2}} + \sqrt{\frac{1}{2}} = \frac{\sqrt{5} + 1}{\sqrt{2}}$.

(B) We are given the expression $\sqrt{12 - \sqrt{68 + 48\sqrt{2}}}$.
First,simplify the inner radical $\sqrt{68 + 48\sqrt{2}}$.
Note that $68 + 48\sqrt{2} = 6^2 + (4\sqrt{2})^2 + 2 \times 6 \times 4\sqrt{2} = (6 + 4\sqrt{2})^2$.
Thus,$\sqrt{68 + 48\sqrt{2}} = 6 + 4\sqrt{2}$.
Substituting this back into the original expression,we get $\sqrt{12 - (6 + 4\sqrt{2})} = \sqrt{6 - 4\sqrt{2}}$.
Now,express $6 - 4\sqrt{2}$ as a perfect square: $6 - 4\sqrt{2} = 2^2 + (\sqrt{2})^2 - 2 \times 2 \times \sqrt{2} = (2 - \sqrt{2})^2$.
Therefore,$\sqrt{6 - 4\sqrt{2}} = \sqrt{(2 - \sqrt{2})^2} = 2 - \sqrt{2}$.

(D) Given expression: $\sqrt{50} + \sqrt{48} = 5\sqrt{2} + 4\sqrt{3}$.
We want to find $\sqrt{\sqrt{50} + \sqrt{48}}$.
Note that $\sqrt{50} + \sqrt{48} = \sqrt{2}(25 + \sqrt{24}) = \sqrt{2}(5^2 + (\sqrt{24})^2 + 2 \times 5 \times \sqrt{24})$ is not quite right. Let us rewrite $\sqrt{50} + \sqrt{48} = 5\sqrt{2} + 4\sqrt{3}$.
Consider the square of $(\sqrt{3} + \sqrt{2})$: $(\sqrt{3} + \sqrt{2})^2 = 3 + 2 + 2\sqrt{6} = 5 + 2\sqrt{6}$.
Then $\sqrt{2}(\sqrt{3} + \sqrt{2})^2 = \sqrt{2}(5 + 2\sqrt{6}) = 5\sqrt{2} + 2\sqrt{12} = 5\sqrt{2} + 4\sqrt{3} = \sqrt{50} + \sqrt{48}$.
Therefore,$\sqrt{\sqrt{50} + \sqrt{48}} = \sqrt{\sqrt{2}(\sqrt{3} + \sqrt{2})^2} = 2^{1/4}(\sqrt{3} + \sqrt{2})$.

(B) To simplify $\sqrt{3 + \sqrt{5}} - \sqrt{2 + \sqrt{3}}$,we multiply and divide by $\sqrt{2}$ inside the square roots:
$\sqrt{3 + \sqrt{5}} - \sqrt{2 + \sqrt{3}} = \sqrt{\frac{6 + 2\sqrt{5}}{2}} - \sqrt{\frac{4 + 2\sqrt{3}}{2}}$
$= \frac{\sqrt{(\sqrt{5})^2 + 1^2 + 2\sqrt{5}}}{\sqrt{2}} - \frac{\sqrt{(\sqrt{3})^2 + 1^2 + 2\sqrt{3}}}{\sqrt{2}}$
$= \frac{\sqrt{(\sqrt{5} + 1)^2}}{\sqrt{2}} - \frac{\sqrt{(\sqrt{3} + 1)^2}}{\sqrt{2}}$
$= \frac{\sqrt{5} + 1}{\sqrt{2}} - \frac{\sqrt{3} + 1}{\sqrt{2}}$
$= \frac{\sqrt{5} + 1 - \sqrt{3} - 1}{\sqrt{2}}$
$= \frac{\sqrt{5} - \sqrt{3}}{\sqrt{2}}$
$= \sqrt{\frac{5}{2}} - \sqrt{\frac{3}{2}}$

(D) Let $x = (9\sqrt{3} + 11\sqrt{2})^{1/3}$.
Then $x^3 = 9\sqrt{3} + 11\sqrt{2}$.
We know that $(a + b)^3 = a^3 + b^3 + 3ab(a + b)$.
Let $a = \sqrt{3}$ and $b = \sqrt{2}$.
Then $a^3 = (\sqrt{3})^3 = 3\sqrt{3}$ and $b^3 = (\sqrt{2})^3 = 2\sqrt{2}$.
$a^3 + b^3 = 3\sqrt{3} + 2\sqrt{2}$.
$3ab(a + b) = 3(\sqrt{3})(\sqrt{2})(\sqrt{3} + \sqrt{2}) = 3\sqrt{6}(\sqrt{3} + \sqrt{2}) = 3(3\sqrt{2} + 2\sqrt{3}) = 9\sqrt{2} + 6\sqrt{3}$.
Adding these,$x^3 = (3\sqrt{3} + 2\sqrt{2}) + (9\sqrt{2} + 6\sqrt{3}) = 9\sqrt{3} + 11\sqrt{2}$.
Thus,$x^3 = (\sqrt{3} + \sqrt{2})^3$.
Therefore,$x = \sqrt{3} + \sqrt{2}$.

(A) First,simplify the numerator term $\sqrt{6 + 2\sqrt{3} + 2\sqrt{2} + 2\sqrt{6}}$.
Note that $(1 + \sqrt{2} + \sqrt{3})^2 = 1^2 + (\sqrt{2})^2 + (\sqrt{3})^2 + 2(1)(\sqrt{2}) + 2(1)(\sqrt{3}) + 2(\sqrt{2})(\sqrt{3}) = 1 + 2 + 3 + 2\sqrt{2} + 2\sqrt{3} + 2\sqrt{6} = 6 + 2\sqrt{2} + 2\sqrt{3} + 2\sqrt{6}$.
Thus,$\sqrt{6 + 2\sqrt{3} + 2\sqrt{2} + 2\sqrt{6}} = 1 + \sqrt{2} + \sqrt{3}$.
Next,simplify the denominator $\sqrt{5 + 2\sqrt{6}}$.
Note that $(\sqrt{3} + \sqrt{2})^2 = 3 + 2 + 2\sqrt{6} = 5 + 2\sqrt{6}$.
Thus,$\sqrt{5 + 2\sqrt{6}} = \sqrt{3} + \sqrt{2}$.
Substituting these into the expression: $\frac{(1 + \sqrt{2} + \sqrt{3}) - 1}{\sqrt{3} + \sqrt{2}} = \frac{\sqrt{3} + \sqrt{2}}{\sqrt{3} + \sqrt{2}} = 1$.

(B) We need to find the square root of $134 + \sqrt{6292}$.
First,simplify $\sqrt{6292} = \sqrt{4 \times 1573} = 2\sqrt{1573} = 2\sqrt{121 \times 13} = 2 \times 11 \sqrt{13} = 22\sqrt{13}$.
Now,the expression is $134 + 22\sqrt{13}$.
We want to write this in the form $(a + b)^2 = a^2 + b^2 + 2ab$.
Let $a = 11$ and $b = \sqrt{13}$.
Then $a^2 + b^2 = 11^2 + (\sqrt{13})^2 = 121 + 13 = 134$.
And $2ab = 2 \times 11 \times \sqrt{13} = 22\sqrt{13}$.
Thus,$134 + 22\sqrt{13} = 11^2 + (\sqrt{13})^2 + 2(11)(\sqrt{13}) = (11 + \sqrt{13})^2$.
Therefore,$\sqrt{134 + \sqrt{6292}} = \sqrt{(11 + \sqrt{13})^2} = 11 + \sqrt{13}$.

(B) Let $\sqrt{-8 - 6i} = x + iy$.
Squaring both sides,we get $-8 - 6i = (x + iy)^2 = x^2 - y^2 + 2xyi$.
Equating real and imaginary parts,we have $x^2 - y^2 = -8$ $(i)$ and $2xy = -6$ $(ii)$.
We know that $(x^2 + y^2)^2 = (x^2 - y^2)^2 + (2xy)^2$.
$(x^2 + y^2)^2 = (-8)^2 + (-6)^2 = 64 + 36 = 100$.
Since $x^2 + y^2$ must be positive,$x^2 + y^2 = 10$ $(iii)$.
Adding $(i)$ and $(iii)$,$2x^2 = 2 \implies x^2 = 1 \implies x = \pm 1$.
Subtracting $(i)$ from $(iii)$,$2y^2 = 18 \implies y^2 = 9 \implies y = \pm 3$.
From $(ii)$,$2xy = -6$,so $x$ and $y$ must have opposite signs.
Thus,the square roots are $\pm(1 - 3i)$.

(B) Given that $\sqrt{-7 - 24i} = x - iy$.
Squaring both sides,we get $-7 - 24i = (x - iy)^2 = x^2 - y^2 - 2xyi$.
Equating the real and imaginary parts,we have $x^2 - y^2 = -7$ and $2xy = 24$.
We know that $(x^2 + y^2)^2 = (x^2 - y^2)^2 + (2xy)^2$.
Substituting the values,$(x^2 + y^2)^2 = (-7)^2 + (24)^2 = 49 + 576 = 625$.
Taking the square root,$x^2 + y^2 = \sqrt{625} = 25$ (since $x^2 + y^2$ must be positive).

(C) Given $\sqrt{x + iy} = \pm(a + ib)$.
Squaring both sides,we get $x + iy = (a + ib)^2 = a^2 - b^2 + 2iab$.
Comparing real and imaginary parts,$x = a^2 - b^2$ and $y = 2ab$.
Now,we need to find $\sqrt{-x - iy} = \sqrt{-(x + iy)}$.
Substituting the values,$\sqrt{-x - iy} = \sqrt{-(a^2 - b^2 + 2iab)} = \sqrt{b^2 - a^2 - 2iab}$.
We can rewrite the expression inside the square root as $(b - ia)^2 = b^2 + (ia)^2 - 2iab = b^2 - a^2 - 2iab$.
Therefore,$\sqrt{-x - iy} = \sqrt{(b - ia)^2} = \pm(b - ia)$.

(A) Let $\sqrt{3 - 4i} = x + iy$.
Squaring both sides,we get $3 - 4i = (x^2 - y^2) + 2ixy$.
Equating real and imaginary parts:
$x^2 - y^2 = 3$ and $2xy = -4$ (so $xy = -2$).
We know that $(x^2 + y^2)^2 = (x^2 - y^2)^2 + (2xy)^2$.
$(x^2 + y^2)^2 = (3)^2 + (-4)^2 = 9 + 16 = 25$.
Thus,$x^2 + y^2 = 5$ (since $x^2 + y^2 > 0$).
Adding $x^2 - y^2 = 3$ and $x^2 + y^2 = 5$,we get $2x^2 = 8$,so $x^2 = 4$,which means $x = \pm 2$.
Subtracting the equations,we get $2y^2 = 2$,so $y^2 = 1$,which means $y = \pm 1$.
Since $xy = -2$ (negative),$x$ and $y$ must have opposite signs.
Therefore,the square roots are $\pm (2 - i)$.

(D) Given $\sqrt{a + ib} = x + iy$.
Squaring both sides,we get $a + ib = (x + iy)^2 = x^2 - y^2 + 2xyi$.
Comparing real and imaginary parts,we have $a = x^2 - y^2$ and $b = 2xy$.
Now,consider $\sqrt{a - ib} = \sqrt{(x^2 - y^2) - 2xyi}$.
This can be written as $\sqrt{x^2 + (iy)^2 - 2x(iy)} = \sqrt{(x - iy)^2} = x - iy$.
Thus,the correct option is $D$.

(A) Let $z = (1 - i)^{-i}$. Taking $\log$ on both sides,
$\log z = -i \log(1 - i) = -i \log \left( \sqrt{2} e^{-i\pi/4} \right)$
$= -i \left[ \log \sqrt{2} + \log(e^{-i\pi/4}) \right]$
$= -i \left[ \frac{1}{2} \log 2 - \frac{i\pi}{4} \right]$
$= -\frac{i}{2} \log 2 - \frac{\pi}{4}$
Therefore,$z = e^{-\pi/4} e^{-i(\frac{1}{2} \log 2)}$.
Using Euler's formula $e^{i\theta} = \cos \theta + i \sin \theta$,we have $z = e^{-\pi/4} \left[ \cos \left( \frac{1}{2} \log 2 \right) - i \sin \left( \frac{1}{2} \log 2 \right) \right]$.
The real part is $\text{Re}(z) = e^{-\pi/4} \cos \left( \frac{1}{2} \log 2 \right)$.

(B) Let $z = i \log \left( \frac{x - i}{x + i} \right)$.
Then $\frac{z}{i} = \log \left( \frac{x - i}{x + i} \right)$.
Multiplying the numerator and denominator by $(x - i)$,we get:
$\frac{z}{i} = \log \left( \frac{(x - i)^2}{x^2 + 1} \right) = \log \left( \frac{x^2 - 1 - 2ix}{x^2 + 1} \right) = \log \left( \frac{x^2 - 1}{x^2 + 1} - i \frac{2x}{x^2 + 1} \right)$.
Using the polar form $\log(a + ib) = \log \sqrt{a^2 + b^2} + i \tan^{-1}(\frac{b}{a})$,where $a = \frac{x^2 - 1}{x^2 + 1}$ and $b = \frac{-2x}{x^2 + 1}$:
$\sqrt{a^2 + b^2} = \sqrt{\frac{(x^2 - 1)^2 + 4x^2}{(x^2 + 1)^2}} = \sqrt{\frac{x^4 + 2x^2 + 1}{(x^2 + 1)^2}} = 1$.
So,$\frac{z}{i} = \log(1) + i \tan^{-1} \left( \frac{-2x / (x^2 + 1)}{(x^2 - 1) / (x^2 + 1)} \right) = 0 + i \tan^{-1} \left( \frac{-2x}{x^2 - 1} \right) = i \tan^{-1} \left( \frac{2x}{1 - x^2} \right)$.
Using the identity $2 \tan^{-1} x = \tan^{-1} \left( \frac{2x}{1 - x^2} \right)$ (for $|x| < 1$),we get $\frac{z}{i} = i(2 \tan^{-1} x)$.
Thus,$z = i^2(2 \tan^{-1} x) = -2 \tan^{-1} x$. Considering the branch of the logarithm,the expression evaluates to $\pi - 2 \tan^{-1} x$.

(A) Let $\sqrt[3]{61 - 46\sqrt{5}} = x - y\sqrt{5}$.
Cubing both sides,we get $61 - 46\sqrt{5} = (x - y\sqrt{5})^3$.
Using the identity $(a - b)^3 = a^3 - 3a^2b + 3ab^2 - b^3$,we have:
$61 - 46\sqrt{5} = x^3 - 3x^2(y\sqrt{5}) + 3x(y\sqrt{5})^2 - (y\sqrt{5})^3$.
$61 - 46\sqrt{5} = x^3 - 3x^2y\sqrt{5} + 15xy^2 - 5y^3\sqrt{5}$.
$61 - 46\sqrt{5} = (x^3 + 15xy^2) - (3x^2y + 5y^3)\sqrt{5}$.
Comparing the rational and irrational parts:
$x^3 + 15xy^2 = 61$ and $3x^2y + 5y^3 = 46$.
If we test $x = 1$ and $y = 2$:
$1^3 + 15(1)(2^2) = 1 + 60 = 61$ (Satisfied).
$3(1^2)(2) + 5(2^3) = 6 + 40 = 46$ (Satisfied).
Thus,$\sqrt[3]{61 - 46\sqrt{5}} = 1 - 2\sqrt{5}$.

(A) We have $\sqrt{17 + 12\sqrt{2}} = \sqrt{3^2 + (2\sqrt{2})^2 + 2 \times 3 \times 2\sqrt{2}}$.
This simplifies to $\sqrt{(3 + 2\sqrt{2})^2} = 3 + 2\sqrt{2}$.
Therefore,$\sqrt[4]{17 + 12\sqrt{2}} = \sqrt{3 + 2\sqrt{2}}$.
Since $3 + 2\sqrt{2} = 2 + 1 + 2\sqrt{2} = (\sqrt{2})^2 + 1^2 + 2\sqrt{2} = (\sqrt{2} + 1)^2$,
we get $\sqrt{3 + 2\sqrt{2}} = \sqrt{(\sqrt{2} + 1)^2} = \sqrt{2} + 1$.

(B) First,simplify the inner nested radical $\sqrt{68 + 48\sqrt{2}}$.
We can write $48\sqrt{2} = 2 \times 24 \times \sqrt{2} = 2 \times \sqrt{576 \times 2} = 2 \times \sqrt{1152}$.
Alternatively,express $68 + 48\sqrt{2}$ in the form $(a + b\sqrt{2})^2 = a^2 + 2b^2 + 2ab\sqrt{2}$.
Comparing,$2ab = 48 \implies ab = 24$. Let $a = 6, b = 4$. Then $a^2 + 2b^2 = 36 + 2(16) = 36 + 32 = 68$.
So,$\sqrt{68 + 48\sqrt{2}} = \sqrt{(6 + 4\sqrt{2})^2} = 6 + 4\sqrt{2}$.
Now,substitute this into the original expression: $\sqrt{12 - (6 + 4\sqrt{2})} = \sqrt{6 - 4\sqrt{2}}$.
We need to simplify $\sqrt{6 - 4\sqrt{2}}$. Let $\sqrt{6 - 4\sqrt{2}} = \sqrt{x} - \sqrt{y}$.
Squaring both sides: $6 - 4\sqrt{2} = x + y - 2\sqrt{xy}$.
Thus,$x + y = 6$ and $2\sqrt{xy} = 4\sqrt{2} \implies \sqrt{xy} = 2\sqrt{2} = \sqrt{8} \implies xy = 8$.
The numbers $x$ and $y$ are $4$ and $2$.
Therefore,$\sqrt{6 - 4\sqrt{2}} = \sqrt{4} - \sqrt{2} = 2 - \sqrt{2}$.

(D) We need to find the square root of $\sqrt{50} + \sqrt{48}$.
First,simplify the expression: $\sqrt{50} + \sqrt{48} = 5\sqrt{2} + 4\sqrt{3}$.
Let $\sqrt{5\sqrt{2} + 4\sqrt{3}} = \sqrt{x} + \sqrt{y}$.
Squaring both sides,we get $5\sqrt{2} + 4\sqrt{3} = x + y + 2\sqrt{xy}$.
This does not simplify easily into the form $\sqrt{a} + \sqrt{b}$.
Let us factor out $2^{1/2}$ from the expression: $\sqrt{50} + \sqrt{48} = \sqrt{2}(5 + \sqrt{24}) = \sqrt{2}(5 + 2\sqrt{6})$.
Now,we need the square root of $\sqrt{2}(5 + 2\sqrt{6})$.
This is equal to $2^{1/4} \times \sqrt{5 + 2\sqrt{6}}$.
We know that $5 + 2\sqrt{6} = 3 + 2 + 2\sqrt{3 \times 2} = (\sqrt{3} + \sqrt{2})^2$.
Therefore,$\sqrt{5 + 2\sqrt{6}} = \sqrt{3} + \sqrt{2}$.
Thus,the expression becomes $2^{1/4}(\sqrt{3} + \sqrt{2})$.

(C) Let the expression be $x = \sqrt{12\sqrt{5} + 2\sqrt{55}}$.
We can rewrite the expression inside the square root as $12\sqrt{5} + 2\sqrt{55} = 2\sqrt{5}(6 + \sqrt{11})$.
This does not immediately simplify to a perfect square of the form $(\sqrt{a} + \sqrt{b})^2$.
Let us re-examine the expression: $\sqrt{12\sqrt{5} + 2\sqrt{55}} = \sqrt{2\sqrt{5} \cdot 6 + 2\sqrt{55}} = \sqrt{2\sqrt{5}(6 + \sqrt{11})}$.
Alternatively,consider the form $(\sqrt{x} + \sqrt{y})^2 = x + y + 2\sqrt{xy}$.
We have $12\sqrt{5} + 2\sqrt{55} = 2\sqrt{5} \cdot 6 + 2\sqrt{55} = 2\sqrt{5} \cdot \sqrt{36} + 2\sqrt{55} = 2\sqrt{180} + 2\sqrt{55}$.
This is not standard. Let's try to express $12\sqrt{5} + 2\sqrt{55}$ as $(\sqrt{a} + \sqrt{b})^2 = a + b + 2\sqrt{ab}$.
$a + b = 12\sqrt{5}$ and $ab = 55$.
Solving $t^2 - (12\sqrt{5})t + 55 = 0$ using the quadratic formula:
$t = \frac{12\sqrt{5} \pm \sqrt{(12\sqrt{5})^2 - 4(55)}}{2} = \frac{12\sqrt{5} \pm \sqrt{720 - 220}}{2} = \frac{12\sqrt{5} \pm \sqrt{500}}{2} = \frac{12\sqrt{5} \pm 10\sqrt{5}}{2}$.
$t_1 = \frac{22\sqrt{5}}{2} = 11\sqrt{5}$ and $t_2 = \frac{2\sqrt{5}}{2} = \sqrt{5}$.
Thus,$12\sqrt{5} + 2\sqrt{55} = (\sqrt{11\sqrt{5}} + \sqrt{\sqrt{5}})^2$.
Taking the square root,we get $\sqrt{11\sqrt{5}} + \sqrt{\sqrt{5}} = \sqrt{11} \cdot 5^{1/4} + 5^{1/4} = 5^{1/4}(\sqrt{11} + 1)$.

(D) Let the cube root be $(a\sqrt{3} + b\sqrt{2})$.
Then,$(a\sqrt{3} + b\sqrt{2})^3 = 9\sqrt{3} + 11\sqrt{2}$.
Expanding the left side: $(a\sqrt{3})^3 + 3(a\sqrt{3})^2(b\sqrt{2}) + 3(a\sqrt{3})(b\sqrt{2})^2 + (b\sqrt{2})^3 = 9\sqrt{3} + 11\sqrt{2}$.
$27a^3\sqrt{3} + 3(3a^2)(b\sqrt{2}) + 3(a\sqrt{3})(2b^2) + 2b^3\sqrt{2} = 9\sqrt{3} + 11\sqrt{2}$.
$9a^3\sqrt{3} + 9a^2b\sqrt{2} + 6ab^2\sqrt{3} + 2b^3\sqrt{2} = 9\sqrt{3} + 11\sqrt{2}$.
Grouping terms with $\sqrt{3}$ and $\sqrt{2}$:
$(9a^3 + 6ab^2)\sqrt{3} + (9a^2b + 2b^3)\sqrt{2} = 9\sqrt{3} + 11\sqrt{2}$.
Comparing coefficients:
$9a^3 + 6ab^2 = 9 \implies 3a^3 + 2ab^2 = 3$.
$9a^2b + 2b^3 = 11$.
If $a=1, b=1$: $3(1)^3 + 2(1)(1)^2 = 3+2 = 5 \neq 3$.
If $a=1, b=1$ for the second equation: $9(1)^2(1) + 2(1)^3 = 9+2 = 11$. This matches.
Let's recheck the first equation with $a=1, b=1$: $9(1)^3 + 6(1)(1)^2 = 9+6 = 15 \neq 9$.
Try $a=1, b=1$ was wrong. Let's test option $B$: $(\sqrt{3} + 2\sqrt{2})^3 = (\sqrt{3})^3 + 3(\sqrt{3})^2(2\sqrt{2}) + 3(\sqrt{3})(2\sqrt{2})^2 + (2\sqrt{2})^3 = 3\sqrt{3} + 3(3)(2\sqrt{2}) + 3\sqrt{3}(8) + 16\sqrt{2} = 3\sqrt{3} + 18\sqrt{2} + 24\sqrt{3} + 16\sqrt{2} = 27\sqrt{3} + 34\sqrt{2}$.
Try option $A$: $(2\sqrt{3} + \sqrt{2})^3 = (2\sqrt{3})^3 + 3(2\sqrt{3})^2(\sqrt{2}) + 3(2\sqrt{3})(\sqrt{2})^2 + (\sqrt{2})^3 = 24\sqrt{3} + 3(12)(\sqrt{2}) + 6\sqrt{3}(2) + 2\sqrt{2} = 24\sqrt{3} + 36\sqrt{2} + 12\sqrt{3} + 2\sqrt{2} = 36\sqrt{3} + 38\sqrt{2}$.
Wait,checking the question again: $(a\sqrt{3} + b\sqrt{2})^3 = a^3(3\sqrt{3}) + 3a^2b(3\sqrt{2}) + 3ab^2(2\sqrt{3}) + b^3(2\sqrt{2}) = (3a^3 + 6ab^2)\sqrt{3} + (9a^2b + 2b^3)\sqrt{2}$.
For $a=1, b=1$: $(3+6)\sqrt{3} + (9+2)\sqrt{2} = 9\sqrt{3} + 11\sqrt{2}$.
Thus,$a=1, b=1$ works,so the cube root is $1\sqrt{3} + 1\sqrt{2} = \sqrt{3} + \sqrt{2}$.

(C) We know that $i = e^{i\pi/2}$.
Substituting this into the expression for $z$:
$z = (e^{i\pi/2})^{2i} = e^{(i\pi/2) \times (2i)} = e^{i^2 \pi} = e^{-\pi}$.
Since $e^{-\pi}$ is a real positive number,its modulus is $|z| = |e^{-\pi}| = e^{-\pi}$.

(N/A) Let $1=r \cos \theta$ and $\sqrt{3}=r \sin \theta$.
By squaring and adding,we get:
$r^{2}(\cos ^{2} \theta+\sin ^{2} \theta)=1^{2}+(\sqrt{3})^{2}$
$r^{2}(1)=1+3=4$
$r=\sqrt{4}=2$ (conventionally,$r>0$).
Therefore,$\cos \theta=\frac{1}{2}$ and $\sin \theta=\frac{\sqrt{3}}{2}$,which gives $\theta=\frac{\pi}{3}$.
Thus,the required polar form is $z=2(\cos \frac{\pi}{3}+i \sin \frac{\pi}{3})$.
The complex number $z=1+i \sqrt{3}$ is represented in the Cartesian plane as shown in the figure.

(A) Let $z = (3+2 \sqrt{-54})^{1/2} - (3-2 \sqrt{-54})^{1/2}$.
First,simplify the expression inside the square root: $\sqrt{-54} = \sqrt{54}i = 3\sqrt{6}i$.
So,$3+2\sqrt{-54} = 3 + 2(3\sqrt{6}i) = 3 + 6\sqrt{6}i$. This does not form a perfect square of the form $(a+bi)^2$ easily.
Let $(3+2\sqrt{-54})^{1/2} = x+iy$. Then $x^2-y^2 = 3$ and $2xy = 2\sqrt{54} = 6\sqrt{6}$.
$x^2+y^2 = \sqrt{3^2 + (6\sqrt{6})^2} = \sqrt{9 + 216} = \sqrt{225} = 15$.
Adding the equations: $2x^2 = 18 \implies x^2 = 9 \implies x = \pm 3$.
Subtracting: $2y^2 = 12 \implies y^2 = 6 \implies y = \pm \sqrt{6}$.
Thus,$(3+2\sqrt{-54})^{1/2} = \pm(3+\sqrt{6}i)$ and $(3-2\sqrt{-54})^{1/2} = \pm(3-\sqrt{6}i)$.
Possible values for the expression are:
$1) (3+\sqrt{6}i) - (3-\sqrt{6}i) = 2\sqrt{6}i$
$2) (3+\sqrt{6}i) - (-(3-\sqrt{6}i)) = 6$
$3) -(3+\sqrt{6}i) - (3-\sqrt{6}i) = -6$
$4) -(3+\sqrt{6}i) - (-(3-\sqrt{6}i)) = -2\sqrt{6}i$.
The imaginary parts are $2\sqrt{6}$ and $-2\sqrt{6}$.
Comparing with options,the correct choice is $-2\sqrt{6}$.

(A) Let $\sqrt{-5-12i} = a+ib$,where $a, b \in \mathbb{R}$.
Squaring both sides,we get $(a+ib)^2 = -5-12i$.
$(a^2-b^2) + i(2ab) = -5-12i$.
Equating real and imaginary parts: $a^2-b^2 = -5$ and $2ab = -12$,which implies $ab = -6$.
Since $b = -6/a$,substituting into the first equation gives $a^2 - (-6/a)^2 = -5$.
$a^2 - 36/a^2 = -5 \implies a^4 + 5a^2 - 36 = 0$.
Factoring the quadratic in $a^2$: $(a^2+9)(a^2-4) = 0$.
Since $a \in \mathbb{R}$,$a^2 = 4$,so $a = \pm 2$.
If $a = 2$,$b = -6/2 = -3$. If $a = -2$,$b = -6/(-2) = 3$.
Thus,the square roots are $\pm(2-3i)$.

(C) Let $z = 2^{-i}$.
Taking the natural logarithm on both sides,we get:
$\log z = \log (2^{-i})$
$\Rightarrow \log z = -i \log 2$
Since $z = e^{\log z}$,we have:
$z = e^{-i \log 2} = e^{i \log(1/2)}$
Using Euler's formula,$e^{i \theta} = \cos \theta + i \sin \theta$,where $\theta = \log(1/2)$:
$z = \cos(\log(1/2)) + i \sin(\log(1/2))$
The real part of $z$ is $\cos(\log(1/2))$.

(A) Let $z = i^{i}$.
Taking the natural logarithm on both sides,we get $\ln(z) = i \ln(i)$.
We know that $i = e^{i(\pi/2 + 2n\pi)}$ for any integer $n$. Taking the principal value where $n=0$,we have $\ln(i) = i\pi/2$.
Substituting this into the equation,we get $\ln(z) = i(i\pi/2) = i^{2}(\pi/2) = -\pi/2$.
Therefore,$z = e^{-\pi/2}$.
Since $e^{-\pi/2}$ is a purely real number,it can be written as $e^{-\pi/2} + 0i$.
Thus,the imaginary part of $i^{i}$ is $0$.

(A) Let $z = \frac{\sqrt{21+12 \sqrt{2} i}}{\sqrt{21-12 \sqrt{2} i}}$.
Note that $21+12 \sqrt{2} i = (2 \sqrt{2} + 3i)^2$ because $(2 \sqrt{2})^2 - 3^2 + 2(2 \sqrt{2})(3)i = 8 - 9 + 12 \sqrt{2} i = -1 + 12 \sqrt{2} i$. This is not correct.
Let us find the square root of $21+12 \sqrt{2} i$. Let $(x+iy)^2 = 21+12 \sqrt{2} i$.
$x^2-y^2 = 21$ and $2xy = 12 \sqrt{2} \implies xy = 6 \sqrt{2}$.
$(x^2+y^2)^2 = (x^2-y^2)^2 + (2xy)^2 = 21^2 + (12 \sqrt{2})^2 = 441 + 288 = 729$.
$x^2+y^2 = 27$.
Adding $x^2-y^2=21$ and $x^2+y^2=27$,we get $2x^2 = 48 \implies x^2 = 24 \implies x = 2 \sqrt{6}$.
$2y^2 = 6 \implies y^2 = 3 \implies y = \sqrt{3}$.
So,$\sqrt{21+12 \sqrt{2} i} = 2 \sqrt{6} + i \sqrt{3}$.
Similarly,$\sqrt{21-12 \sqrt{2} i} = 2 \sqrt{6} - i \sqrt{3}$.
Then $z = \frac{2 \sqrt{6} + i \sqrt{3}}{2 \sqrt{6} - i \sqrt{3}} = \frac{(2 \sqrt{6} + i \sqrt{3})^2}{(2 \sqrt{6})^2 + (\sqrt{3})^2} = \frac{24 - 3 + 4 \sqrt{18} i}{24 + 3} = \frac{21 + 12 \sqrt{2} i}{27} = \frac{21}{27} + i \frac{12 \sqrt{2}}{27} = \frac{7}{9} + i \frac{4 \sqrt{2}}{9}$.
Here $a = \frac{7}{9}$ and $b = \frac{4 \sqrt{2}}{9}$.
Thus,$\frac{b}{a} = \frac{4 \sqrt{2} / 9}{7 / 9} = \frac{4 \sqrt{2}}{7}$.

(C) Let $\sqrt{-5-12 i} = x+y i$,where $x > 0$.
$-5-12 i = (x+y i)^2 = x^2-y^2+2 x y i$.
Equating real and imaginary parts: $x^2-y^2 = -5$ and $2 x y = -12$.
$x^2+y^2 = \sqrt{(-5)^2+(-12)^2} = \sqrt{25+144} = 13$.
Adding the equations $x^2-y^2 = -5$ and $x^2+y^2 = 13$ gives $2 x^2 = 8$,so $x = 2$ (since $x > 0$).
Then $y = -3$. Thus,$\sqrt{-5-12 i} = 2-3 i$.
Similarly,for $\sqrt{5+12 i} = u+v i$ with $u > 0$: $u^2-v^2 = 5$ and $2 u v = 12$.
$u^2+v^2 = 13$. Adding gives $2 u^2 = 18$,so $u = 3$ and $v = 2$. Thus,$\sqrt{5+12 i} = 3+2 i$.
For $\sqrt{-8-6 i} = m+n i$ with $m < 0$: $m^2-n^2 = -8$ and $2 m n = -6$.
$m^2+n^2 = \sqrt{(-8)^2+(-6)^2} = 10$.
Adding gives $2 m^2 = 2$,so $m = -1$ (since $m < 0$).
Then $n = 3$. Thus,$\sqrt{-8-6 i} = -1+3 i$.
Now,$a+b i = \frac{(2-3 i)+(3+2 i)}{-1+3 i} = \frac{5-i}{-1+3 i}$.
Multiplying by the conjugate: $\frac{5-i}{-1+3 i} \times \frac{-1-3 i}{-1-3 i} = \frac{-5-15 i+i+3 i^2}{1+9} = \frac{-8-14 i}{10} = -0.8-1.4 i$.
So $a = -0.8$ and $b = -1.4$.
$2 a+b = 2(-0.8)+(-1.4) = -1.6-1.4 = -3$.

(A) Let $\sqrt{7+24 i} = x+iy$. Squaring both sides,we get $7+24 i = (x^2-y^2) + 2xyi$.
Equating the real and imaginary parts,we have $x^2-y^2 = 7$ and $2xy = 24$,which implies $xy = 12$ or $y = \frac{12}{x}$.
Substituting $y$ into the first equation: $x^2 - (\frac{12}{x})^2 = 7 \Rightarrow x^2 - \frac{144}{x^2} = 7$.
Multiplying by $x^2$: $x^4 - 7x^2 - 144 = 0$.
Let $u = x^2$,then $u^2 - 7u - 144 = 0$. Factoring gives $(u-16)(u+9) = 0$.
Since $x$ is real,$x^2 = 16$,so $x = \pm 4$.
If $x = 4$,$y = 3$. If $x = -4$,$y = -3$.
Thus,the square roots are $\pm(4+3i)$.

(A) Given the inequality: $\log_{\cos^2 \theta} |z - a| > \log_{\cos^2 \theta} |z - ai|$.
Since $\cos^2 \theta$ is the base of the logarithm,we must have $\cos^2 \theta \in (0, 1)$.
Because the base is between $0$ and $1$,the inequality reverses when removing the logarithm:
$|z - a| < |z - ai|$.
Substituting $z = x + iy$:
$|x + iy - a| < |x + iy - ai|$
$|(x - a) + iy| < |x + i(y - a)|$.
Squaring both sides:
$(x - a)^2 + y^2 < x^2 + (y - a)^2$
$x^2 - 2ax + a^2 + y^2 < x^2 + y^2 - 2ay + a^2$.
Subtracting $x^2 + y^2 + a^2$ from both sides:
$-2ax < -2ay$.
Since $a > 0$,dividing by $-2a$ reverses the inequality:
$x > y$.

(A) Given the equation: $2 \cosh 2x + 10 \sinh 2x = 5$.
Using the definitions $\cosh 2x = \frac{e^{2x} + e^{-2x}}{2}$ and $\sinh 2x = \frac{e^{2x} - e^{-2x}}{2}$:
$2 \left( \frac{e^{2x} + e^{-2x}}{2} \right) + 10 \left( \frac{e^{2x} - e^{-2x}}{2} \right) = 5$
$(e^{2x} + e^{-2x}) + 5(e^{2x} - e^{-2x}) = 5$
$6e^{2x} - 4e^{-2x} = 5$
Multiply by $e^{2x}$: $6(e^{2x})^2 - 5e^{2x} - 4 = 0$.
Let $u = e^{2x}$,then $6u^2 - 5u - 4 = 0$.
Factoring the quadratic: $(3u - 4)(2u + 1) = 0$.
So,$u = \frac{4}{3}$ or $u = -\frac{1}{2}$.
Since $e^{2x} > 0$,we have $e^{2x} = \frac{4}{3}$.
Taking the natural logarithm on both sides: $2x = \ln \frac{4}{3}$.
Therefore,$x = \frac{1}{2} \ln \frac{4}{3}$.

(C) We use the logarithmic forms of inverse hyperbolic functions:
$\sec h^{-1} x = \log _e\left(\frac{1+\sqrt{1-x^2}}{x}\right)$ and $\operatorname{cosec} h^{-1} x = \log _e\left(\frac{1+\sqrt{1+x^2}}{x}\right)$.
For $\sec h^{-1}\left(\frac{1}{2}\right)$:
$\sec h^{-1}\left(\frac{1}{2}\right) = \log _e\left(\frac{1+\sqrt{1-(1/2)^2}}{1/2}\right) = \log _e\left(\frac{1+\sqrt{3/4}}{1/2}\right) = \log _e\left(\frac{1+\sqrt{3}/2}{1/2}\right) = \log _e(2+\sqrt{3})$.
For $\operatorname{cosec} h^{-1}\left(\frac{3}{4}\right)$:
$\operatorname{cosec} h^{-1}\left(\frac{3}{4}\right) = \log _e\left(\frac{1+\sqrt{1+(3/4)^2}}{3/4}\right) = \log _e\left(\frac{1+\sqrt{25/16}}{3/4}\right) = \log _e\left(\frac{1+5/4}{3/4}\right) = \log _e\left(\frac{9/4}{3/4}\right) = \log _e(3)$.
Subtracting the two values:
$\log _e(2+\sqrt{3}) - \log _e(3) = \log _e\left(\frac{2+\sqrt{3}}{3}\right)$.

(B) We know that $\cosh(x) = \frac{e^x + e^{-x}}{2}$.
Substituting $x = \log 4$,we get $\cosh(\log 4) = \frac{e^{\log 4} + e^{-\log 4}}{2}$.
Since $e^{\log 4} = 4$ and $e^{-\log 4} = e^{\log(4^{-1})} = \frac{1}{4}$,
$\cosh(\log 4) = \frac{4 + \frac{1}{4}}{2} = \frac{\frac{17}{4}}{2} = \frac{17}{8}$.

(C) We know the relationship between the inverse hyperbolic tangent and the inverse trigonometric tangent function: $\tanh^{-1}(z) = \frac{1}{i} \tan^{-1}(iz)$.
To find $\tanh^{-1}(iy)$,we substitute $z = iy$ into the formula:
$\tanh^{-1}(iy) = \frac{1}{i} \tan^{-1}(i(iy))$
Since $i^2 = -1$,this simplifies to:
$\tanh^{-1}(iy) = \frac{1}{i} \tan^{-1}(-y)$
Using the property $\tan^{-1}(-y) = -\tan^{-1}(y)$,we get:
$\tanh^{-1}(iy) = \frac{1}{i} (-\tan^{-1}(y)) = -\frac{1}{i} \tan^{-1}(y)$
Since $\frac{1}{i} = -i$,we have $-\frac{1}{i} = i$.
Therefore,$\tanh^{-1}(iy) = i \tan^{-1}(y)$.
Thus,option $C$ is correct.

(D) Let $z_1 = 24-70 i$ and $z_2 = -24+70 i$.
We need to find $\sqrt{z_1} + \sqrt{z_2}$.
Note that $z_2 = -z_1$,so we are looking for $\sqrt{z_1} + \sqrt{-z_1}$.
Let $\sqrt{24-70 i} = x+iy$. Squaring both sides,$x^2-y^2+2ixy = 24-70 i$.
Equating real and imaginary parts: $x^2-y^2 = 24$ and $2xy = -70 \implies xy = -35$.
Since $x^2+y^2 = \sqrt{24^2+(-70)^2} = \sqrt{576+4900} = \sqrt{5476} = 74$.
Adding $x^2-y^2=24$ and $x^2+y^2=74$,we get $2x^2 = 98 \implies x^2 = 49 \implies x = \pm 7$.
If $x=7, y=-5$. If $x=-7, y=5$. So $\sqrt{24-70 i} = \pm(7-5 i)$.
Similarly,$\sqrt{-24+70 i} = \pm(5-7 i)$.
Possible sums are:
$1$) $(7-5 i) + (5-7 i) = 12-12 i$
$2$) $(7-5 i) - (5-7 i) = 2+2 i$
$3$) $-(7-5 i) + (5-7 i) = -2-2 i$
$4$) $-(7-5 i) - (5-7 i) = -12+12 i$
Wait,checking the options provided,there might be a typo in the question or options. Re-evaluating: $\sqrt{24-70 i} = \pm(7-5 i)$ and $\sqrt{-24+70 i} = \pm i(7-5 i) = \pm(5+7 i)$.
Sum: $\pm(7-5 i) \pm(5+7 i)$.
Possible values: $(7-5 i) + (5+7 i) = 12+2 i$,$(7-5 i) - (5+7 i) = 2-12 i$,$-(7-5 i) + (5+7 i) = -2+12 i$,$-(7-5 i) - (5+7 i) = -12-2 i$.
Option $D$ matches $-12-2 i$.

(C) Let $Z = \sqrt{-5-12 i} + \sqrt{7+24 i}$.
We use the formula $\sqrt{a+ib} = \pm \left( \sqrt{\frac{|Z|+a}{2}} + i \frac{b}{|b|} \sqrt{\frac{|Z|-a}{2}} \right)$.
For $\sqrt{-5-12 i}$,$|Z| = \sqrt{(-5)^2 + (-12)^2} = 13$. Thus,$\sqrt{-5-12 i} = \pm \left( \sqrt{\frac{13-5}{2}} - i \sqrt{\frac{13+5}{2}} \right) = \pm(2-3i)$.
For $\sqrt{7+24 i}$,$|Z| = \sqrt{7^2 + 24^2} = 25$. Thus,$\sqrt{7+24 i} = \pm \left( \sqrt{\frac{25+7}{2}} + i \sqrt{\frac{25-7}{2}} \right) = \pm(4+3i)$.
Given $Z = \pm(2-3i) \pm(4+3i)$.
Possible values for $Z$ are $(2-3i) + (4+3i) = 6$,$(2-3i) - (4+3i) = -2-6i$,$-(2-3i) + (4+3i) = 2+6i$,and $-(2-3i) - (4+3i) = -6$.
Since $k$ is a negative real number,$k = -6$.

(C) Let $z = \sqrt{(-3+4 i)(8+6 i)}$.
First,calculate the product inside the square root:
$(-3+4 i)(8+6 i) = -24 - 18 i + 32 i + 24 i^2$
Since $i^2 = -1$,we have:
$-24 + 14 i - 24 = -48 + 14 i$.
Now,we need to find $\sqrt{-48 + 14 i}$.
Let $\sqrt{-48 + 14 i} = x + i y$,where $x, y \in \mathbb{R}$.
Squaring both sides: $(x + i y)^2 = -48 + 14 i$
$x^2 - y^2 + 2 i x y = -48 + 14 i$.
Comparing real and imaginary parts:
$x^2 - y^2 = -48$ $(1)$
$2 x y = 14 \Rightarrow x y = 7$ $(2)$
From $(2)$,$y = \frac{7}{x}$. Substituting into $(1)$:
$x^2 - (\frac{7}{x})^2 = -48$
$x^2 - \frac{49}{x^2} = -48$
$x^4 + 48 x^2 - 49 = 0$
$(x^2 + 49)(x^2 - 1) = 0$.
Since $x \in \mathbb{R}$,$x^2 = 1$,so $x = \pm 1$.
If $x = 1$,$y = 7$. If $x = -1$,$y = -7$.
Thus,the square roots are $\pm(1 + 7 i)$.

(B) We know that $\sinh(x) = \frac{e^x - e^{-x}}{2}$.
Let $x = \log(3+\sqrt{8})$.
Then $e^x = 3+\sqrt{8} = 3+2\sqrt{2} = (\sqrt{2}+1)^2$.
Also,$e^{-x} = \frac{1}{e^x} = \frac{1}{(\sqrt{2}+1)^2} = (\sqrt{2}-1)^2$.
Substituting these into the formula:
$\sinh(x) = \frac{(\sqrt{2}+1)^2 - (\sqrt{2}-1)^2}{2}$.
Expanding the squares:
$(\sqrt{2}+1)^2 = 2+1+2\sqrt{2} = 3+2\sqrt{2}$.
$(\sqrt{2}-1)^2 = 2+1-2\sqrt{2} = 3-2\sqrt{2}$.
$\sinh(x) = \frac{(3+2\sqrt{2}) - (3-2\sqrt{2})}{2} = \frac{4\sqrt{2}}{2} = 2\sqrt{2}$.
Since $2\sqrt{2} = 2^1 \times 2^{1/2} = 2^{3/2}$,the correct option is $B$.

(A) Given: $x = \log \left(\cot \left(\frac{\pi}{4} + \theta\right)\right)$
Substitute $\theta = \frac{\pi}{12}$:
$x = \log \left(\cot \left(\frac{\pi}{4} + \frac{\pi}{12}\right)\right) = \log \left(\cot \left(\frac{3\pi + \pi}{12}\right)\right) = \log \left(\cot \left(\frac{\pi}{3}\right)\right)$
Since $\cot \left(\frac{\pi}{3}\right) = \frac{1}{\sqrt{3}}$,we have $x = \log \left(\frac{1}{\sqrt{3}}\right) = -\log \sqrt{3} = -\frac{1}{2} \log 3$.
Then $e^x = \frac{1}{\sqrt{3}}$ and $e^{-x} = \sqrt{3}$.
The definition of $\cosh x$ is $\frac{e^x + e^{-x}}{2}$.
$\cosh x = \frac{\frac{1}{\sqrt{3}} + \sqrt{3}}{2} = \frac{\frac{1 + 3}{\sqrt{3}}}{2} = \frac{4}{2\sqrt{3}} = \frac{2}{\sqrt{3}}$.

(B) Given $\sinh x = \frac{3}{4}$.
Using the formula $\sinh^{-1} z = \log(z + \sqrt{z^2 + 1})$:
$x = \log(\frac{3}{4} + \sqrt{(\frac{3}{4})^2 + 1}) = \log(\frac{3}{4} + \sqrt{\frac{9}{16} + 1}) = \log(\frac{3}{4} + \frac{5}{4}) = \log(2)$.
Given $\cosh y = \frac{5}{3}$.
Using the formula $\cosh^{-1} z = \log(z + \sqrt{z^2 - 1})$:
$y = \log(\frac{5}{3} + \sqrt{(\frac{5}{3})^2 - 1}) = \log(\frac{5}{3} + \sqrt{\frac{25}{9} - 1}) = \log(\frac{5}{3} + \frac{4}{3}) = \log(3)$.
Therefore,$x + y = \log 2 + \log 3 = \log(2 \times 3) = \log 6$.

(D) We have,$\cosh ^{-1} x = 2 \log _e(\sqrt{2}+1)$.
Using the logarithmic property $n \log a = \log a^n$,we get:
$\cosh ^{-1} x = \log _e(\sqrt{2}+1)^2$.
We know that $\cosh ^{-1} x = \log _e(x + \sqrt{x^2-1})$.
Therefore,$\log _e(x + \sqrt{x^2-1}) = \log _e(2 + 1 + 2\sqrt{2}) = \log _e(3 + 2\sqrt{2})$.
Comparing both sides,we have $x + \sqrt{x^2-1} = 3 + 2\sqrt{2}$.
Let $x = 3$,then $\sqrt{x^2-1} = \sqrt{9-1} = \sqrt{8} = 2\sqrt{2}$.
Thus,$3 + 2\sqrt{2} = 3 + 2\sqrt{2}$,which satisfies the equation.
Hence,$x = 3$.

(A) We use the logarithmic definitions of inverse hyperbolic functions:
$\sinh^{-1}(x) = \log(x + \sqrt{x^2+1})$
$\operatorname{cosech}^{-1}(x) = \log\left(\frac{1}{x} + \sqrt{\frac{1}{x^2}+1}\right)$
$\coth^{-1}(x) = \frac{1}{2} \log\left(\frac{x+1}{x-1}\right)$
Substituting $x = -2$:
$\sinh^{-1}(-2) = \log(-2 + \sqrt{5})$
$\operatorname{cosech}^{-1}(-2) = \log\left(-\frac{1}{2} + \sqrt{\frac{1}{4}+1}\right) = \log\left(\frac{-1+\sqrt{5}}{2}\right)$
$\coth^{-1}(-2) = \frac{1}{2} \log\left(\frac{-2+1}{-2-1}\right) = \frac{1}{2} \log\left(\frac{-1}{-3}\right) = \frac{1}{2} \log\left(\frac{1}{3}\right) = \log\left(\frac{1}{\sqrt{3}}\right)$
Summing these:
$\log(-2 + \sqrt{5}) + \log\left(\frac{\sqrt{5}-1}{2}\right) + \log\left(\frac{1}{\sqrt{3}}\right) = \log\left[(-2 + \sqrt{5}) \cdot \left(\frac{\sqrt{5}-1}{2}\right) \cdot \frac{1}{\sqrt{3}}\right]$
$= \log\left[\frac{-2\sqrt{5} + 2 + 5 - \sqrt{5}}{2\sqrt{3}}\right] = \log\left(\frac{7-3\sqrt{5}}{2\sqrt{3}}\right)$

(A) We know that the formula for the inverse hyperbolic sine function is $\sinh^{-1}(x) = \log\{x + \sqrt{x^2 + 1}\}$.
Substituting $x = 2^{3/2}$ into the formula:
$\sinh^{-1}(2^{3/2}) = \log\{2^{3/2} + \sqrt{(2^{3/2})^2 + 1\}}$.
Since $2^{3/2} = \sqrt{2^3} = \sqrt{8}$,we have:
$\sinh^{-1}(2^{3/2}) = \log\{\sqrt{8} + \sqrt{8 + 1\}}$.
$= \log\{\sqrt{8} + \sqrt{9\}}$.
$= \log\{3 + \sqrt{8\}}$.
Thus,the correct option is $A$.

(D) We use the logarithmic definitions of inverse hyperbolic functions:
$\sinh^{-1}(x) = \ln(x + \sqrt{x^2+1})$
$\cosh^{-1}(x) = \ln(x + \sqrt{x^2-1})$
$\tanh^{-1}(x) = \frac{1}{2} \ln\left(\frac{1+x}{1-x}\right)$
$\coth^{-1}(x) = \frac{1}{2} \ln\left(\frac{x+1}{x-1}\right)$
Substituting the values:
$\sinh^{-1}(2) = \ln(2 + \sqrt{5})$
$\cosh^{-1}(2) = \ln(2 + \sqrt{3})$
$\tanh^{-1}\left(\frac{2}{3}\right) = \frac{1}{2} \ln\left(\frac{1+2/3}{1-2/3}\right) = \frac{1}{2} \ln(5)$
$\coth^{-1}(-2) = \frac{1}{2} \ln\left(\frac{-2+1}{-2-1}\right) = \frac{1}{2} \ln\left(\frac{-1}{-3}\right) = \frac{1}{2} \ln\left(\frac{1}{3}\right) = -\frac{1}{2} \ln(3)$
Now,sum them up:
$\ln(2+\sqrt{5}) + \ln(2+\sqrt{3}) - \frac{1}{2} \ln(5) - \frac{1}{2} \ln(3)$
$= \ln((2+\sqrt{5})(2+\sqrt{3})) - \frac{1}{2} \ln(15)$
$= \ln((2+\sqrt{5})(2+\sqrt{3})) - \ln(\sqrt{15})$
$= \ln\left(\frac{(2+\sqrt{5})(2+\sqrt{3})}{\sqrt{15}}\right)$
$= \ln\left(\frac{(2+\sqrt{5})(2+\sqrt{3}) \sqrt{3}}{\sqrt{5} \cdot 3} \cdot \sqrt{3}\right) = \ln\left(\frac{(2+\sqrt{5})(2+\sqrt{3}) \sqrt{3}}{\sqrt{5}}\right)$
Thus,the correct option is $D$.

(C) We need to evaluate $\operatorname{sech}^{-1}\left(\frac{3}{5}\right)-\tanh ^{-1}\left(\frac{3}{5}\right)$.
First,we use the identity $\operatorname{sech}^{-1}(x) = \cosh^{-1}\left(\frac{1}{x}\right)$.
So,$\operatorname{sech}^{-1}\left(\frac{3}{5}\right) = \cosh^{-1}\left(\frac{5}{3}\right)$.
Using the logarithmic form $\cosh^{-1}(x) = \log_e(x + \sqrt{x^2 - 1})$:
$\cosh^{-1}\left(\frac{5}{3}\right) = \log_e\left(\frac{5}{3} + \sqrt{\frac{25}{9} - 1}\right) = \log_e\left(\frac{5}{3} + \sqrt{\frac{16}{9}}\right) = \log_e\left(\frac{5}{3} + \frac{4}{3}\right) = \log_e(3)$.
Next,we use the identity $\tanh^{-1}(x) = \frac{1}{2} \log_e\left(\frac{1+x}{1-x}\right)$.
So,$\tanh^{-1}\left(\frac{3}{5}\right) = \frac{1}{2} \log_e\left(\frac{1 + 3/5}{1 - 3/5}\right) = \frac{1}{2} \log_e\left(\frac{8/5}{2/5}\right) = \frac{1}{2} \log_e(4) = \frac{1}{2} \log_e(2^2) = \log_e(2)$.
Finally,subtracting the two values:
$\operatorname{sech}^{-1}\left(\frac{3}{5}\right) - \tanh^{-1}\left(\frac{3}{5}\right) = \log_e(3) - \log_e(2) = \log_e\left(\frac{3}{2}\right)$.

(C) Given expression: $\sqrt{12-\sqrt{68+48 \sqrt{2}}}$
First,simplify the inner radical: $\sqrt{68+48 \sqrt{2}} = \sqrt{68+2 \times 24 \sqrt{2}} = \sqrt{68+2 \times 6 \times 4 \sqrt{2}}$
Since $(6)^2 + (4 \sqrt{2})^2 = 36 + 32 = 68$,we have $\sqrt{68+48 \sqrt{2}} = \sqrt{(6+4 \sqrt{2})^2} = 6+4 \sqrt{2}$
Substitute this back: $\sqrt{12-(6+4 \sqrt{2})} = \sqrt{6-4 \sqrt{2}}$
Now,express $6-4 \sqrt{2}$ as a perfect square: $6-4 \sqrt{2} = 4 + 2 - 2 \times 2 \times \sqrt{2} = (2)^2 + (\sqrt{2})^2 - 2 \times 2 \times \sqrt{2} = (2-\sqrt{2})^2$
Therefore,$\sqrt{6-4 \sqrt{2}} = \sqrt{(2-\sqrt{2})^2} = 2-\sqrt{2}$