Integral power of iota, Algebraic operations and Equality of complex numbers Questions in English

(B) Since the coefficients are real,if $1 + i$ is a root,then its conjugate $1 - i$ must also be a root.
The sum of the roots is $(1 + i) + (1 - i) = 2$.
The product of the roots is $(1 + i)(1 - i) = 1^2 - i^2 = 1 - (-1) = 2$.
The quadratic equation is given by $x^2 - (\text{sum of roots})x + (\text{product of roots}) = 0$.
Substituting the values,we get $x^2 - 2x + 2 = 0$.

(B) Let $x$ be a non-zero rational number and $y$ be an irrational number.
Assume that the product $xy = r$,where $r$ is a rational number.
Since $x \neq 0$,we can write $y = \frac{r}{x}$.
Since $r$ is rational and $x$ is a non-zero rational,the quotient of two rational numbers is always a rational number.
This implies that $y$ must be a rational number,which contradicts the given condition that $y$ is an irrational number.
Therefore,the assumption that $xy$ is rational is false.
Hence,$xy$ must be an irrational number.
For example,if $x = 2$ and $y = \sqrt{3}$,then $xy = 2\sqrt{3}$,which is irrational.

(B) For a complex number to be purely imaginary,its real part must be equal to $0$.
Given expression: $Z = \frac{2 + 3i \sin \theta}{1 - 2i \sin \theta}$
Multiply the numerator and denominator by the conjugate of the denominator $(1 + 2i \sin \theta)$:
$Z = \frac{(2 + 3i \sin \theta)(1 + 2i \sin \theta)}{(1 - 2i \sin \theta)(1 + 2i \sin \theta)}$
$Z = \frac{2 + 4i \sin \theta + 3i \sin \theta + 6i^2 \sin^2 \theta}{1 - (2i \sin \theta)^2}$
Since $i^2 = -1$:
$Z = \frac{2 + 7i \sin \theta - 6 \sin^2 \theta}{1 + 4 \sin^2 \theta}$
$Z = \frac{2 - 6 \sin^2 \theta}{1 + 4 \sin^2 \theta} + i \frac{7 \sin \theta}{1 + 4 \sin^2 \theta}$
Setting the real part to $0$:
$\frac{2 - 6 \sin^2 \theta}{1 + 4 \sin^2 \theta} = 0$
$2 - 6 \sin^2 \theta = 0$
$6 \sin^2 \theta = 2$
$\sin^2 \theta = \frac{1}{3}$
$\sin \theta = \pm \frac{1}{\sqrt{3}}$
Therefore,$\theta = \sin^{-1}\left(\frac{1}{\sqrt{3}}\right)$.

(A) Given the equation: ${\left( {\frac{{1 + i}}{{1 - i}}} \right)^x} = 1$.
First,simplify the term inside the bracket by multiplying the numerator and denominator by the conjugate of the denominator $(1 + i)$:
$\frac{{1 + i}}{{1 - i}} \times \frac{{1 + i}}{{1 + i}} = \frac{{(1 + i)^2}}{{1^2 - i^2}} = \frac{{1 + i^2 + 2i}}{{1 - (-1)}} = \frac{{1 - 1 + 2i}}{2} = \frac{{2i}}{2} = i$.
Substituting this back into the equation,we get: $i^x = 1$.
We know that $i^k = 1$ if and only if $k$ is a multiple of $4$.
Therefore,$x = 4n$,where $n$ is any positive integer.

(C) Given that $|a| = a$ and $|b| = b$,this implies $a \ge 0$ and $b \ge 0$.
We need to evaluate the expression $\left( \frac{a}{a^2} - \frac{b}{b^2} \right)^2$.
First,simplify the terms inside the parenthesis:
$\frac{a}{a^2} = \frac{1}{a}$ and $\frac{b}{b^2} = \frac{1}{b}$.
Substituting these into the expression,we get:
$\left( \frac{1}{a} - \frac{1}{b} \right)^2 = \left( \frac{b - a}{ab} \right)^2$.
Since $(b - a)^2 = (a - b)^2$,we have:
$\left( \frac{a - b}{ab} \right)^2$.

(B) Given: $x + \sqrt{x^2 + 1} = a$
Let $x + \sqrt{x^2 + 1} = a$ --- $(1)$
We know that $(\sqrt{x^2 + 1} + x)(\sqrt{x^2 + 1} - x) = (x^2 + 1) - x^2 = 1$
So,$a(\sqrt{x^2 + 1} - x) = 1$
$\sqrt{x^2 + 1} - x = \frac{1}{a}$ --- $(2)$
Subtracting equation $(2)$ from equation $(1)$:
$(x + \sqrt{x^2 + 1}) - (\sqrt{x^2 + 1} - x) = a - \frac{1}{a}$
$2x = a - \frac{1}{a}$
$x = \frac{1}{2}(a - \frac{1}{a})$

(A) The given sum is $S = \sum\limits_{n = 1}^{50} {{i^{2n - 1}}} = i^1 + i^3 + i^5 + i^7 + \dots + i^{99}$.
This is a geometric progression with first term $a = i$,common ratio $r = i^2 = -1$,and number of terms $n = 50$.
The sum of a geometric progression is given by $S = \frac{a(1 - r^n)}{1 - r}$.
Substituting the values: $S = \frac{i(1 - (-1)^{50})}{1 - (-1)} = \frac{i(1 - 1)}{2} = \frac{i(0)}{2} = 0$.

(C) The given sum is $S = \sum_{n=1}^{50} i^{(2n-1)!} = i^{1!} + i^{3!} + i^{5!} + i^{7!} + \dots + i^{99!}$.
For $n=1$,$i^{1!} = i^1 = i$.
For $n=2$,$i^{3!} = i^6 = i^4 \times i^2 = 1 \times (-1) = -1$.
For $n \ge 3$,$(2n-1)!$ is a multiple of $4$ (since $(2n-1)!$ contains $4 \times 3 \times 2 \times 1$ as factors for $n \ge 3$).
Thus,$i^{(2n-1)!} = (i^4)^k = 1^k = 1$ for all $n \ge 3$.
There are $50 - 3 + 1 = 48$ terms from $n=3$ to $n=50$,each equal to $1$.
Therefore,$S = i + (-1) + 48(1) = i - 1 + 48 = 47 + i$.

(D) Let $z = \frac{3 + 2i \sin \theta}{1 - 2i \sin \theta}$.
Multiplying the numerator and denominator by the conjugate of the denominator:
$z = \frac{(3 + 2i \sin \theta)(1 + 2i \sin \theta)}{(1 - 2i \sin \theta)(1 + 2i \sin \theta)}$
$z = \frac{3 + 6i \sin \theta + 2i \sin \theta - 4 \sin^2 \theta}{1 + 4 \sin^2 \theta} = \frac{(3 - 4 \sin^2 \theta) + 8i \sin \theta}{1 + 4 \sin^2 \theta}$.
For $z$ to be purely imaginary,the real part must be zero:
$3 - 4 \sin^2 \theta = 0 \implies \sin^2 \theta = \frac{3}{4} \implies \sin \theta = \pm \frac{\sqrt{3}}{2}$.
Given $\theta \in \left( -\frac{\pi}{2}, \pi \right)$:
If $\sin \theta = \frac{\sqrt{3}}{2}$,then $\theta = \frac{\pi}{3}$ or $\theta = \frac{2\pi}{3}$.
If $\sin \theta = -\frac{\sqrt{3}}{2}$,then $\theta = -\frac{\pi}{3}$.
The set $A = \left\{ -\frac{\pi}{3}, \frac{\pi}{3}, \frac{2\pi}{3} \right\}$.
The sum of the elements is $-\frac{\pi}{3} + \frac{\pi}{3} + \frac{2\pi}{3} = \frac{2\pi}{3}$.

(A) Given: ${\left( -2 - \frac{i}{3} \right)^3} = \frac{x + iy}{27}$
Take out the negative sign: $-1 \times {\left( 2 + \frac{i}{3} \right)^3} = \frac{x + iy}{27}$
Expand using $(a + b)^3 = a^3 + b^3 + 3a^2b + 3ab^2$:
$-1 \times \left[ 2^3 + \left( \frac{i}{3} \right)^3 + 3(2^2)\left( \frac{i}{3} \right) + 3(2)\left( \frac{i}{3} \right)^2 \right] = \frac{x + iy}{27}$
$-1 \times \left[ 8 - \frac{i}{27} + 4i - \frac{2}{3} \right] = \frac{x + iy}{27}$
$-8 + \frac{2}{3} - i\left( 4 - \frac{1}{27} \right) = \frac{x + iy}{27}$
Multiply by $27$:
$x + iy = 27 \times \left( -\frac{22}{3} \right) - i \times 27 \times \left( \frac{107}{27} \right)$
$x = -198$ and $y = -107$
Then $y - x = -107 - (-198) = -107 + 198 = 91$

(C) Let $z = x + 10i$.
Given $\frac{2z - n}{2z + n} = 2i - 1$.
Substituting $z = x + 10i$:
$\frac{2(x + 10i) - n}{2(x + 10i) + n} = 2i - 1$
$(2x - n) + 20i = (2i - 1)((2x + n) + 20i)$
$(2x - n) + 20i = 2i(2x + n) + 40i^2 - (2x + n) - 20i$
$(2x - n) + 20i = (2i(2x + n) - 20i) - 40 - (2x + n)$
Equating real and imaginary parts:
Real part: $2x - n = -40 - (2x + n)$ $\Rightarrow$ $2x - n = -40 - 2x - n$ $\Rightarrow$ $4x = -40$ $\Rightarrow$ $x = -10$.
Imaginary part: $20 = 2(2x + n) - 20$ $\Rightarrow$ $40 = 2(2x + n)$ $\Rightarrow$ $20 = 2x + n$.
Substituting $x = -10$ into $20 = 2x + n$:
$20 = 2(-10) + n$ $\Rightarrow$ $20 = -20 + n$ $\Rightarrow$ $n = 40$.
Thus, $n = 40$ and $Re(z) = -10$.

(A) Given the equation: $4x + i(3x - y) = 3 + i(-6)$.
Equating the real and imaginary parts on both sides,we get:
Real part: $4x = 3 \implies x = \frac{3}{4}$.
Imaginary part: $3x - y = -6$.
Substitute $x = \frac{3}{4}$ into the imaginary part equation:
$3(\frac{3}{4}) - y = -6$
$\frac{9}{4} - y = -6$
$y = \frac{9}{4} + 6$
$y = \frac{9 + 24}{4} = \frac{33}{4}$.
Thus,$x = \frac{3}{4}$ and $y = \frac{33}{4}$.

(A) Given expression: $(-5i) \left(\frac{1}{8}i\right)$
Multiply the coefficients and the imaginary units:
$= \left(-5 \times \frac{1}{8}\right) \times (i \times i)$
$= -\frac{5}{8} \times i^2$
Since $i^2 = -1$,substitute this value:
$= -\frac{5}{8} \times (-1)$
$= \frac{5}{8}$
Expressing in the form $a+bi$:
$= \frac{5}{8} + i0$

(A) Given expression: $(-i)(2i)\left(-\frac{1}{8}i\right)^{3}$
$= (-2i^{2}) \times \left(-\frac{1}{512}i^{3}\right)$
$= (-2(-1)) \times \left(-\frac{1}{512}(-i)\right)$
$= 2 \times \left(\frac{1}{512}i\right)$
$= \frac{1}{256}i$
Thus,the expression in $a+bi$ form is $0 + \frac{1}{256}i$.

(B) We use the identity $(x-y)^3 = x^3 - 3x^2y + 3xy^2 - y^3$.
Here,$x = 5$ and $y = 3i$.
$(5-3i)^3 = 5^3 - 3(5^2)(3i) + 3(5)(3i)^2 - (3i)^3$
$= 125 - 3(25)(3i) + 15(9i^2) - 27i^3$
$= 125 - 225i + 135(-1) - 27(-i)$
$= 125 - 225i - 135 + 27i$
$= (125 - 135) + (-225 + 27)i$
$= -10 - 198i$

(A) Given expression: $(-\sqrt{3}+\sqrt{-2})(2 \sqrt{3}-i)$
Since $\sqrt{-2} = \sqrt{2}i$,the expression becomes $(-\sqrt{3}+\sqrt{2}i)(2\sqrt{3}-i)$.
Expanding the product:
$= (-\sqrt{3})(2\sqrt{3}) - (-\sqrt{3})(i) + (\sqrt{2}i)(2\sqrt{3}) - (\sqrt{2}i)(i)$
$= -6 + \sqrt{3}i + 2\sqrt{6}i - \sqrt{2}i^2$
Since $i^2 = -1$,we have:
$= -6 + \sqrt{3}i + 2\sqrt{6}i + \sqrt{2}$
$= (-6+\sqrt{2}) + i(\sqrt{3} + 2\sqrt{6})$
$= (-6+\sqrt{2}) + i\sqrt{3}(1 + 2\sqrt{2})$

(B) To express the complex number in the form $a+ib$,we multiply the numerator and the denominator by the conjugate of the denominator:
$\frac{5+\sqrt{2}i}{1-\sqrt{2}i} = \frac{5+\sqrt{2}i}{1-\sqrt{2}i} \times \frac{1+\sqrt{2}i}{1+\sqrt{2}i}$
$= \frac{5(1) + 5(\sqrt{2}i) + (\sqrt{2}i)(1) + (\sqrt{2}i)(\sqrt{2}i)}{1^2 - (\sqrt{2}i)^2}$
$= \frac{5 + 5\sqrt{2}i + \sqrt{2}i + 2i^2}{1 - 2i^2}$
Since $i^2 = -1$,we have:
$= \frac{5 + 6\sqrt{2}i - 2}{1 - 2(-1)} = \frac{3 + 6\sqrt{2}i}{1 + 2}$
$= \frac{3 + 6\sqrt{2}i}{3} = 1 + 2\sqrt{2}i$

(A) Given expression is $i^{-35}$.
We know that $i^{-35} = \frac{1}{i^{35}}$.
Since $i^4 = 1$,we can write $i^{35} = i^{32} \times i^3 = (i^4)^8 \times i^3 = 1^8 \times (-i) = -i$.
Therefore,$i^{-35} = \frac{1}{-i}$.
Multiplying the numerator and denominator by $i$,we get $\frac{1 \times i}{-i \times i} = \frac{i}{-i^2} = \frac{i}{-(-1)} = \frac{i}{1} = i$.
In the form $a+ib$,this is $0+1i$.

(B) Given expression: $(5i)\left(-\frac{3}{5}i\right)$
$= 5 \times \left(-\frac{3}{5}\right) \times i \times i$
$= -3 \times i^{2}$
Since $i^{2} = -1$,we have:
$= -3(-1)$
$= 3$
To express in the form $a+ib$,we write it as $3+0i$.

(A) We know that $i^{4} = 1$.
$i^{9} = i^{4 \times 2 + 1} = (i^{4})^{2} \times i = (1)^{2} \times i = i$.
$i^{19} = i^{4 \times 4 + 3} = (i^{4})^{4} \times i^{3} = (1)^{4} \times (-i) = -i$.
Therefore,$i^{9} + i^{19} = i + (-i) = 0$.
In the form $a+ib$,this is $0 + 0i$.

(A) Given expression: $i^{-39}$
We know that $i^4 = 1$. We can write $-39$ as $4 \times (-10) + 1$.
$i^{-39} = \frac{1}{i^{39}}$
$= \frac{1}{i^{4 \times 9 + 3}} = \frac{1}{(i^4)^9 \cdot i^3}$
$= \frac{1}{(1)^9 \cdot (-i)} = \frac{1}{-i}$
Multiply numerator and denominator by $i$:
$= \frac{1 \cdot i}{-i \cdot i} = \frac{i}{-i^2} = \frac{i}{-(-1)} = \frac{i}{1} = i$
In the form $a+ib$,this is $0+i$.

(A) Given expression: $3(7+i7)+i(7+i7)$
Expand the terms:
$= 3 \times 7 + 3 \times i7 + i \times 7 + i \times i7$
$= 21 + 21i + 7i + 7i^2$
Combine the imaginary parts:
$= 21 + 28i + 7i^2$
Substitute $i^2 = -1$:
$= 21 + 28i + 7(-1)$
$= 21 + 28i - 7$
$= 14 + 28i$

(A) To express the complex number $(1-i)-(-1+i6)$ in the form $a+ib$,we perform the subtraction:
$(1-i)-(-1+i6) = 1-i+1-6i$
Grouping the real and imaginary parts:
$= (1+1) + (-i-6i)$
$= 2-7i$

(A) Given expression: $\left(\frac{1}{5}+i \frac{2}{5}\right)-\left(4+i \frac{5}{2}\right)$
$= \frac{1}{5} + i \frac{2}{5} - 4 - i \frac{5}{2}$
$= \left(\frac{1}{5} - 4\right) + i \left(\frac{2}{5} - \frac{5}{2}\right)$
$= \left(\frac{1-20}{5}\right) + i \left(\frac{4-25}{10}\right)$
$= \frac{-19}{5} + i \left(\frac{-21}{10}\right)$
$= \frac{-19}{5} - \frac{21}{10}i$

(A) Given expression: $\left[\left(\frac{1}{3}+i \frac{7}{3}\right)+\left(4+i \frac{1}{3}\right)\right]-\left(-\frac{4}{3}+i\right)$
$= \left(\frac{1}{3}+4\right) + i\left(\frac{7}{3}+\frac{1}{3}\right) + \frac{4}{3} - i$
$= \left(\frac{1}{3}+4+\frac{4}{3}\right) + i\left(\frac{7}{3}+\frac{1}{3}-1\right)$
$= \left(\frac{1+12+4}{3}\right) + i\left(\frac{7+1-3}{3}\right)$
$= \frac{17}{3} + i \frac{5}{3}$

(A) We have $(1-i)^{4} = [(1-i)^{2}]^{2}$.
Expanding $(1-i)^{2}$ using the identity $(a-b)^{2} = a^{2} + b^{2} - 2ab$:
$(1-i)^{2} = 1^{2} + i^{2} - 2i = 1 - 1 - 2i = -2i$.
Now,substitute this back into the expression:
$(1-i)^{4} = (-2i)^{2}$.
Calculate the square:
$(-2i)^{2} = (-2)^{2} \times i^{2} = 4 \times (-1) = -4$.
Thus,the complex number in the form $a+ib$ is $-4 + 0i$.

(A) Using the identity $(x+y)^{3} = x^{3} + y^{3} + 3xy(x+y)$:
$\left(\frac{1}{3}+3i\right)^{3} = \left(\frac{1}{3}\right)^{3} + (3i)^{3} + 3\left(\frac{1}{3}\right)(3i)\left(\frac{1}{3}+3i\right)$
$= \frac{1}{27} + 27i^{3} + 3i\left(\frac{1}{3}+3i\right)$
$= \frac{1}{27} - 27i + i + 9i^{2} \quad [\text{since } i^{3} = -i, i^{2} = -1]$
$= \frac{1}{27} - 27i + i - 9$
$= \left(\frac{1}{27} - 9\right) + i(-27 + 1)$
$= \frac{1-243}{27} - 26i$
$= \frac{-242}{27} - 26i$

(A) We use the identity $(x+y)^{3} = x^{3} + y^{3} + 3xy(x+y)$.
Given expression: $\left(-2-\frac{1}{3}i\right)^{3} = -\left(2+\frac{1}{3}i\right)^{3}$
$= -\left[2^{3} + \left(\frac{1}{3}i\right)^{3} + 3(2)\left(\frac{1}{3}i\right)\left(2+\frac{1}{3}i\right)\right]$
$= -\left[8 - \frac{1}{27}i + 2i\left(2+\frac{1}{3}i\right)\right]$
$= -\left[8 - \frac{1}{27}i + 4i + \frac{2}{3}i^{2}\right]$
Since $i^{2} = -1$,we have:
$= -\left[8 - \frac{1}{27}i + 4i - \frac{2}{3}\right]$
$= -\left[\left(8-\frac{2}{3}\right) + \left(4-\frac{1}{27}\right)i\right]$
$= -\left[\frac{22}{3} + \frac{107}{27}i\right]$
$= -\frac{22}{3} - \frac{107}{27}i$

(A) Let $z = -i$.
The multiplicative inverse of a complex number $z$ is given by $z^{-1} = \frac{1}{z}$.
$z^{-1} = \frac{1}{-i}$.
Multiply the numerator and denominator by $i$:
$z^{-1} = \frac{1 \times i}{-i \times i} = \frac{i}{-i^2}$.
Since $i^2 = -1$,we have:
$z^{-1} = \frac{i}{-(-1)} = \frac{i}{1} = i$.

(A) Given expression: $\frac{(3+i\sqrt{5})(3-i\sqrt{5})}{(\sqrt{3}+\sqrt{2}i)-(\sqrt{3}-i\sqrt{2})}$
Using the identity $(a+b)(a-b) = a^2-b^2$ in the numerator:
$= \frac{3^2 - (i\sqrt{5})^2}{\sqrt{3} + \sqrt{2}i - \sqrt{3} + i\sqrt{2}}$
$= \frac{9 - 5i^2}{2\sqrt{2}i}$
Since $i^2 = -1$:
$= \frac{9 - 5(-1)}{2\sqrt{2}i} = \frac{14}{2\sqrt{2}i} = \frac{7}{\sqrt{2}i}$
Multiply numerator and denominator by $i$:
$= \frac{7i}{\sqrt{2}i^2} = \frac{7i}{\sqrt{2}(-1)} = \frac{-7i}{\sqrt{2}}$
Rationalizing the denominator:
$= \frac{-7i \times \sqrt{2}}{\sqrt{2} \times \sqrt{2}} = \frac{-7\sqrt{2}i}{2}$

(B) Given equation is $x^{2}+2=0$.
$x^{2} = -2$.
Taking the square root on both sides,we get $x = \pm \sqrt{-2}$.
Since $\sqrt{-1} = i$,we have $x = \pm \sqrt{2} i$.

(A) The given quadratic equation is $x^{2}+3=0$.
Subtracting $3$ from both sides,we get $x^{2}=-3$.
Taking the square root on both sides,we have $x = \pm \sqrt{-3}$.
Since $\sqrt{-1} = i$,we can write this as $x = \pm \sqrt{3} i$.

(A) We have the expression $\left[i^{18}+\left(\frac{1}{i}\right)^{25}\right]^{3}$.
First,simplify $i^{18}$ and $\left(\frac{1}{i}\right)^{25}$:
$i^{18} = (i^4)^4 \cdot i^2 = 1^4 \cdot (-1) = -1$.
$\left(\frac{1}{i}\right)^{25} = \frac{1}{i^{25}} = \frac{1}{(i^4)^6 \cdot i} = \frac{1}{1^6 \cdot i} = \frac{1}{i} = \frac{i}{i^2} = \frac{i}{-1} = -i$.
Substitute these values back into the expression:
$[-1 + (-i)]^3 = [-(1+i)]^3 = (-1)^3 (1+i)^3$.
Expand $(1+i)^3$ using the formula $(a+b)^3 = a^3 + b^3 + 3ab(a+b)$:
$(1+i)^3 = 1^3 + i^3 + 3(1)(i)(1+i) = 1 - i + 3i(1+i) = 1 - i + 3i + 3i^2 = 1 + 2i - 3 = -2 + 2i$.
Finally,multiply by $(-1)^3 = -1$:
$-1 \cdot (-2 + 2i) = 2 - 2i$.

Let $z_{1}=x_{1}+i y_{1}$ and $z_{2}=x_{2}+i y_{2}.$
Then,$z_{1} z_{2}=(x_{1}+i y_{1})(x_{2}+i y_{2}).$
Expanding the product,we get $z_{1} z_{2}=x_{1} x_{2}+i x_{1} y_{2}+i y_{1} x_{2}+i^{2} y_{1} y_{2}.$
Since $i^{2}=-1,$ we have $z_{1} z_{2}=x_{1} x_{2}+i x_{1} y_{2}+i y_{1} x_{2}-y_{1} y_{2}.$
Grouping the real and imaginary parts,$z_{1} z_{2}=(x_{1} x_{2}-y_{1} y_{2})+i(x_{1} y_{2}+y_{1} x_{2}).$
The real part is $\operatorname{Re}(z_{1} z_{2})=x_{1} x_{2}-y_{1} y_{2}.$
Since $\operatorname{Re} z_{1}=x_{1}, \operatorname{Re} z_{2}=x_{2}, \operatorname{Im} z_{1}=y_{1},$ and $\operatorname{Im} z_{2}=y_{2},$
we get $\operatorname{Re}(z_{1} z_{2})=\operatorname{Re} z_{1} \operatorname{Re} z_{2}-\operatorname{Im} z_{1} \operatorname{Im} z_{2}.$
Hence,proved.

(A) First,simplify the expression inside the first bracket: $\frac{1}{1-4 i}-\frac{2}{1+i} = \frac{(1+i)-2(1-4 i)}{(1-4 i)(1+i)} = \frac{1+i-2+8 i}{1+i-4 i-4 i^{2}} = \frac{-1+9 i}{5-3 i}$.
Next,multiply this result by the second term: $\left(\frac{-1+9 i}{5-3 i}\right)\left(\frac{3-4 i}{5+i}\right) = \frac{(-1+9 i)(3-4 i)}{(5-3 i)(5+i)}$.
Expand the numerator: $(-1)(3) + (-1)(-4 i) + (9 i)(3) + (9 i)(-4 i) = -3 + 4 i + 27 i - 36 i^{2} = -3 + 31 i + 36 = 33 + 31 i$.
Expand the denominator: $(5)(5) + (5)(i) + (-3 i)(5) + (-3 i)(i) = 25 + 5 i - 15 i - 3 i^{2} = 25 - 10 i + 3 = 28 - 10 i$.
Now,we have $\frac{33+31 i}{28-10 i}$. To simplify,multiply the numerator and denominator by the conjugate of the denominator,$28+10 i$: $\frac{(33+31 i)(28+10 i)}{(28-10 i)(28+10 i)} = \frac{924 + 330 i + 868 i + 310 i^{2}}{28^{2} + 10^{2}} = \frac{924 + 1198 i - 310}{784 + 100} = \frac{614 + 1198 i}{884}$.
Divide both parts by $2$: $\frac{307}{442} + \frac{599}{442} i$.

(N/A) Given $a+ib = \frac{(x+i)^{2}}{2x^{2}+1}$.
Expanding the numerator using $(A+B)^{2} = A^{2}+B^{2}+2AB$:
$a+ib = \frac{x^{2}+i^{2}+2xi}{2x^{2}+1}$
Since $i^{2} = -1$:
$a+ib = \frac{x^{2}-1+i(2x)}{2x^{2}+1}$
Separating the real and imaginary parts:
$a+ib = \frac{x^{2}-1}{2x^{2}+1} + i\left(\frac{2x}{2x^{2}+1}\right)$
Comparing real and imaginary parts:
$a = \frac{x^{2}-1}{2x^{2}+1}$ and $b = \frac{2x}{2x^{2}+1}$
Now,calculate $a^{2}+b^{2}$:
$a^{2}+b^{2} = \left(\frac{x^{2}-1}{2x^{2}+1}\right)^{2} + \left(\frac{2x}{2x^{2}+1}\right)^{2}$
$= \frac{(x^{2}-1)^{2} + (2x)^{2}}{(2x^{2}+1)^{2}}$
$= \frac{x^{4}+1-2x^{2}+4x^{2}}{(2x^{2}+1)^{2}}$
$= \frac{x^{4}+2x^{2}+1}{(2x^{2}+1)^{2}}$
$= \frac{(x^{2}+1)^{2}}{(2x^{2}+1)^{2}}$
Thus,$a^{2}+b^{2} = \frac{(x^{2}+1)^{2}}{(2x^{2}+1)^{2}}$. Hence,proved.

(D) Given $\left(\frac{1+i}{1-i}\right)^{m}=1$.
First,simplify the expression inside the parenthesis by multiplying the numerator and denominator by the conjugate of the denominator:
$\frac{1+i}{1-i} \times \frac{1+i}{1+i} = \frac{(1+i)^2}{1^2 - i^2} = \frac{1 + i^2 + 2i}{1 - (-1)} = \frac{1 - 1 + 2i}{2} = \frac{2i}{2} = i$.
Substituting this back into the equation,we get $i^m = 1$.
We know that $i^n = 1$ when $n$ is a multiple of $4$.
The powers of $i$ are $i^1 = i, i^2 = -1, i^3 = -i, i^4 = 1$.
Thus,the least positive integral value of $m$ for which $i^m = 1$ is $m = 4$.

(C) Given expression: $E = \frac{(2i)^{n}}{(1-i)^{n-2}}$
First,simplify $(1-i)^2 = 1 + i^2 - 2i = 1 - 1 - 2i = -2i$.
Thus,$(1-i)^{n-2} = ((1-i)^2)^{\frac{n-2}{2}} = (-2i)^{\frac{n-2}{2}}$.
Substituting this into the expression: $E = \frac{(2i)^n}{(-2i)^{\frac{n-2}{2}}} = \frac{(2i)^n}{(-1)^{\frac{n-2}{2}} (2i)^{\frac{n-2}{2}}} = \frac{(2i)^{n - \frac{n-2}{2}}}{(-1)^{\frac{n-2}{2}}} = \frac{(2i)^{\frac{n+2}{2}}}{(-1)^{\frac{n-2}{2}}}$.
For $E$ to be a positive integer,the power of $i$ must be a multiple of $4$ (i.e.,$\frac{n+2}{2} = 4k$) and the magnitude must be real.
If $n=6$,$\frac{n+2}{2} = 4$. Then $E = \frac{(2i)^4}{(-1)^2} = \frac{16 i^4}{1} = 16(1) = 16$,which is a positive integer.
Thus,the least positive integer $n$ is $6$.

(B) Given $Z_1 = 4i^{40} - 5i^{35} + 6i^{17} + 2$.
Since $i^4 = 1$,$i^{40} = (i^4)^{10} = 1$.
$i^{35} = i^{32} \times i^3 = 1 \times (-i) = -i$.
$i^{17} = i^{16} \times i = 1 \times i = i$.
Substituting these values: $Z_1 = 4(1) - 5(-i) + 6(i) + 2 = 4 + 5i + 6i + 2 = 6 + 11i$.
Given $Z_2 = -1 + i$.
$Z_1 + Z_2 = (6 + 11i) + (-1 + i) = (6 - 1) + (11i + i) = 5 + 12i$.
The modulus $|Z_1 + Z_2| = \sqrt{5^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13$.

(D) Given expression: $\frac{i^{248}+i^{246}+i^{244}+i^{242}+i^{240}}{i^{249}+i^{247}+i^{245}+i^{243}+i^{241}}$
Factor out $i^{240}$ from the numerator and $i^{241}$ from the denominator:
$= \frac{i^{240}(i^8+i^6+i^4+i^2+1)}{i^{241}(i^8+i^6+i^4+i^2+1)}$
Cancel the common term $(i^8+i^6+i^4+i^2+1)$:
$= \frac{i^{240}}{i^{241}}$
$= \frac{1}{i}$
Multiply numerator and denominator by $i$:
$= \frac{i}{i^2} = \frac{i}{-1} = -i$

(B) Given expression: $\frac{i^{592}+i^{590}+i^{588}+i^{586}+i^{584}}{i^{582}+i^{580}+i^{578}+i^{576}+i^{574}}-1$
Factor out $i^{584}$ from the numerator and $i^{574}$ from the denominator:
$\frac{i^{584}(i^8+i^6+i^4+i^2+1)}{i^{574}(i^8+i^6+i^4+i^2+1)}-1$
Cancel the common term $(i^8+i^6+i^4+i^2+1)$:
$\frac{i^{584}}{i^{574}}-1 = i^{584-574}-1 = i^{10}-1$
Since $i^4 = 1$,$i^{10} = (i^4)^2 \times i^2 = 1^2 \times (-1) = -1$
Therefore,$-1 - 1 = -2$

(D) Given $\left(-2-\frac{1}{3} i\right)^3=\frac{x+i y}{27}$.
We can rewrite the left side as $\left(\frac{-6-i}{3}\right)^3 = \frac{(-6-i)^3}{27}$.
Equating the numerators,we have $x+iy = (-6-i)^3$.
Using the identity $(a+b)^3 = a^3+3a^2b+3ab^2+b^3$:
$(-6-i)^3 = (-6)^3 + 3(-6)^2(-i) + 3(-6)(-i)^2 + (-i)^3$.
$= -216 + 3(36)(-i) + 3(-6)(-1) - (-i)$.
$= -216 - 108i + 18 + i$.
$= -198 - 107i$.
Comparing with $x+iy$,we get $x = -198$ and $y = -107$.
Therefore,$y-x = -107 - (-198) = -107 + 198 = 91$.

(A) Given $x = \frac{5}{1-2i}$. Multiplying numerator and denominator by the conjugate $(1+2i)$:
$x = \frac{5(1+2i)}{(1-2i)(1+2i)} = \frac{5(1+2i)}{1+4} = \frac{5(1+2i)}{5} = 1+2i$.
Now,calculate $x^2$:
$x^2 = (1+2i)^2 = 1^2 + (2i)^2 + 2(1)(2i) = 1 - 4 + 4i = -3 + 4i$.
Now,calculate $x^3$:
$x^3 = x^2 \cdot x = (-3+4i)(1+2i) = -3 - 6i + 4i + 8i^2 = -3 - 2i - 8 = -11 - 2i$.
Substitute these values into the expression $x^3 + x^2 - x + 22$:
$(-11 - 2i) + (-3 + 4i) - (1 + 2i) + 22$
$= (-11 - 3 - 1 + 22) + (-2i + 4i - 2i)$
$= 7 + 0i = 7$.

(B) Given $\operatorname{Im}(z)=10$, let $z=x+10i$.
The given equation is $\frac{2z-n}{2z+n}=2i-1$.
Substituting $z=x+10i$:
$\frac{2(x+10i)-n}{2(x+10i)+n}=2i-1$
$(2x-n)+20i=(2i-1)(2x+n+20i)$
$(2x-n)+20i = 4xi + 2ni - 40 - 2x - n - 20i$
$(2x-n)+20i = (-2x-n-40) + (4x+2n-20)i$
Equating real and imaginary parts:
Real part: $2x-n = -2x-n-40$ $\Rightarrow 4x = -40$ $\Rightarrow x = -10$.
Imaginary part: $20 = 4x+2n-20$.
Substituting $x=-10$: $20 = 4(-10)+2n-20$ $\Rightarrow 20 = -40+2n-20$ $\Rightarrow 2n = 80$ $\Rightarrow n = 40$.
Thus, $n=40$ and $\operatorname{Re}(z)=x=-10$.

(D) The conjugate of $-6-24i$ is $-6+24i$.
Given that $(x-iy)(3+5i) = -6+24i$.
Expanding the left side: $3x + 5xi - 3yi - 5yi^2 = -6+24i$.
Since $i^2 = -1$,we have $(3x+5y) + (5x-3y)i = -6+24i$.
Equating real and imaginary parts:
$3x + 5y = -6$ (Equation $1$)
$5x - 3y = 24$ (Equation $2$)
Multiplying Equation $1$ by $3$ and Equation $2$ by $5$:
$9x + 15y = -18$
$25x - 15y = 120$
Adding these equations: $34x = 102$,which gives $x = 3$.
Substituting $x=3$ into Equation $1$: $3(3) + 5y = -6$ $\Rightarrow 9 + 5y = -6$ $\Rightarrow 5y = -15$ $\Rightarrow y = -3$.
Thus,$x=3$ and $y=-3$.

(A) First,simplify the expression inside the parenthesis: $\frac{1-i}{1+i} = \frac{(1-i)(1-i)}{(1+i)(1-i)} = \frac{1-2i+i^2}{1-i^2} = \frac{1-2i-1}{1+1} = \frac{-2i}{2} = -i$.
Now,raise this to the power of $100$: $(-i)^{100} = (-1)^{100} \times i^{100} = 1 \times (i^4)^{25} = 1 \times (1)^{25} = 1$.
Given that $a+ib = 1$,we can write this as $1+0i$.
Comparing the real and imaginary parts,we get $a=1$ and $b=0$.
Thus,$(a, b) = (1, 0)$.

(C) Let $z = \frac{2+3i \sin \theta}{1-2i \sin \theta}$.
To simplify, multiply the numerator and denominator by the conjugate of the denominator, $(1+2i \sin \theta)$:
$z = \frac{(2+3i \sin \theta)(1+2i \sin \theta)}{(1-2i \sin \theta)(1+2i \sin \theta)}$
$z = \frac{2 + 4i \sin \theta + 3i \sin \theta + 6i^2 \sin^2 \theta}{1 + 4 \sin^2 \theta}$
Since $i^2 = -1$, we have:
$z = \frac{(2 - 6 \sin^2 \theta) + i(7 \sin \theta)}{1 + 4 \sin^2 \theta}$
For $z$ to be purely imaginary, the real part must be zero:
$\operatorname{Re}(z) = \frac{2 - 6 \sin^2 \theta}{1 + 4 \sin^2 \theta} = 0$
$2 - 6 \sin^2 \theta = 0$
$\sin^2 \theta = \frac{2}{6} = \frac{1}{3}$
$\sin \theta = \frac{1}{\sqrt{3}}$
$\theta = \sin^{-1}\left(\frac{1}{\sqrt{3}}\right)$

(C) Given: $\frac{3+2i}{1+i} = \frac{1}{2}(x+iy)$
$\therefore x+iy = \frac{2(3+2i)}{1+i} \times \frac{1-i}{1-i}$
$= \frac{2(3 - 3i + 2i - 2i^2)}{1 - i^2}$
$= \frac{2(3 - i + 2)}{1 + 1} = \frac{2(5 - i)}{2} = 5 - i$
Comparing real and imaginary parts,we get $x = 5$ and $y = -1$.
Therefore,$x - y = 5 - (-1) = 5 + 1 = 6$.

(D) For a complex number $z$ to be purely real,its imaginary part must be zero.
Given $z = \frac{4 + 3i \sin \theta}{1 - 2i \sin \theta}$.
Multiply the numerator and denominator by the conjugate of the denominator $(1 + 2i \sin \theta)$:
$z = \frac{(4 + 3i \sin \theta)(1 + 2i \sin \theta)}{(1 - 2i \sin \theta)(1 + 2i \sin \theta)}$
$z = \frac{4 + 8i \sin \theta + 3i \sin \theta + 6i^2 \sin^2 \theta}{1 + 4 \sin^2 \theta}$
Since $i^2 = -1$,we have:
$z = \frac{4 - 6 \sin^2 \theta + i(11 \sin \theta)}{1 + 4 \sin^2 \theta}$
For $z$ to be purely real,the imaginary part must be zero:
$\frac{11 \sin \theta}{1 + 4 \sin^2 \theta} = 0$
This implies $11 \sin \theta = 0$,so $\sin \theta = 0$.
The general solution for $\sin \theta = 0$ is $\theta = n \pi$,where $n \in Z$.

(B) Two complex numbers $z_1 = a+bi$ and $z_2 = c+di$ are conjugates if $a=c$ and $b=-d$.
Given $z_1 = (3x+2) - (5y-3)i$ and $z_2 = (6x+3) + (2y-4)i$.
Equating the real parts: $3x+2 = 6x+3$ $\Rightarrow 3x = -1$ $\Rightarrow x = -\frac{1}{3}$.
Equating the imaginary parts: $-(5y-3) = 2y-4$ $\Rightarrow -5y+3 = 2y-4$ $\Rightarrow 7y = 7$ $\Rightarrow y = 1$.
Wait,let's re-evaluate the conjugate condition: $z_1 = \overline{z_2}$.
$(3x+2) - (5y-3)i = (6x+3) - (2y-4)i$.
Real parts: $3x+2 = 6x+3$ $\Rightarrow 3x = -1$ $\Rightarrow x = -\frac{1}{3}$.
Imaginary parts: $-(5y-3) = -(2y-4)$ $\Rightarrow 5y-3 = 2y-4$ $\Rightarrow 3y = -1$ $\Rightarrow y = -\frac{1}{3}$.
Now,calculate $\frac{x-y}{x+y} = \frac{-\frac{1}{3} - (-\frac{1}{3})}{-\frac{1}{3} + (-\frac{1}{3})} = \frac{0}{-\frac{2}{3}} = 0$.