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

(B) Given that,$\left(\frac{1-i}{1+i}\right)^{96}=a+ib$.
First,simplify the expression inside the bracket:
$\frac{1-i}{1+i} = \frac{(1-i)(1-i)}{(1+i)(1-i)} = \frac{1-2i+i^2}{1-i^2} = \frac{1-2i-1}{1+1} = \frac{-2i}{2} = -i$.
Now,substitute this back into the equation:
$(-i)^{96} = a+ib$.
Since $96$ is an even number,$(-i)^{96} = i^{96}$.
$i^{96} = (i^4)^{24} = (1)^{24} = 1$.
So,$1 = a+ib$,which can be written as $1+0i = a+ib$.
Comparing the real and imaginary parts,we get $a=1$ and $b=0$.
Therefore,$(a, b) = (1, 0)$.

(A) Given the expression: $i^{n} + i^{n+1} + i^{n+2} + i^{n+3}$
Factor out $i^{n}$ from the expression: $i^{n}(1 + i + i^{2} + i^{3})$
We know that $i^{2} = -1$ and $i^{3} = -i$.
Substituting these values: $i^{n}(1 + i - 1 - i)$
Simplifying the terms inside the parentheses: $i^{n}(0) = 0$
Therefore,the simplified form is $0$.

(C) Given expression: $\frac{(1+i)^{n}}{(1-i)^{n-2}}$
$= \frac{(1+i)^{n}}{(1-i)^{n}} \times (1-i)^{2}$
$= \left(\frac{1+i}{1-i}\right)^{n} \times (1 + i^{2} - 2i)$
$= \left(\frac{(1+i)(1+i)}{(1-i)(1+i)}\right)^{n} \times (1 - 1 - 2i)$
$= \left(\frac{1 + i^{2} + 2i}{1 - i^{2}}\right)^{n} \times (-2i)$
$= \left(\frac{2i}{2}\right)^{n} \times (-2i)$
$= i^{n} \times (-2i) = -2i^{n+1}$
For the expression to be a positive real number,$-2i^{n+1}$ must be positive.
If $n=1$,$-2i^{1+1} = -2i^{2} = -2(-1) = 2$,which is positive.
Thus,the least positive integer $n$ is $1$.

(C) Let $z = \frac{1-i \sin \alpha}{1+2 i \sin \alpha}$.
To make $z$ purely real,its imaginary part must be zero.
Multiply the numerator and denominator by the conjugate of the denominator $(1-2i \sin \alpha)$:
$z = \frac{(1-i \sin \alpha)(1-2i \sin \alpha)}{(1+2i \sin \alpha)(1-2i \sin \alpha)}$
$z = \frac{1 - 2i \sin \alpha - i \sin \alpha + 2i^2 \sin^2 \alpha}{1 + 4 \sin^2 \alpha}$
Since $i^2 = -1$,we have:
$z = \frac{1 - 3i \sin \alpha - 2 \sin^2 \alpha}{1 + 4 \sin^2 \alpha} = \frac{1 - 2 \sin^2 \alpha}{1 + 4 \sin^2 \alpha} - i \frac{3 \sin \alpha}{1 + 4 \sin^2 \alpha}$
For $z$ to be purely real,the imaginary part must be $0$:
$-\frac{3 \sin \alpha}{1 + 4 \sin^2 \alpha} = 0$
$\Rightarrow 3 \sin \alpha = 0$
$\Rightarrow \sin \alpha = 0$
$\Rightarrow \alpha = n \pi, n \in N$.

(C) Given the equation: $3x + i(4x - y) = 6 - i$
On comparing the real and imaginary parts of the $LHS$ and $RHS$,we get:
$3x = 6$ and $4x - y = -1$
From $3x = 6$,we find $x = 2$.
Substituting $x = 2$ into the equation $4x - y = -1$:
$4(2) - y = -1$
$8 - y = -1$
$y = 8 + 1 = 9$
Therefore,the values are $x = 2$ and $y = 9$.

(D) Given,$\left(\frac{1+i}{1-i}\right)^{x}=1$
Rationalizing the base:
$\left[\frac{1+i}{1-i} \times \frac{1+i}{1+i}\right]^{x}=1$
$\left[\frac{1+i^2+2i}{1^2-i^2}\right]^{x}=1$
Since $i^2 = -1$:
$\left[\frac{1-1+2i}{1+1}\right]^{x}=1$
$\left[\frac{2i}{2}\right]^{x}=1$
$i^x = 1$
We know that $i^k = 1$ if and only if $k$ is a multiple of $4$.
Therefore,$x = 4n$ for $n \in N$.

(C) Given that,$\left(\frac{1+i}{1-i}\right)^{m} = 1$.
First,simplify the expression inside the bracket by multiplying the numerator and denominator by the conjugate of the denominator $(1+i)$:
$\frac{1+i}{1-i} \times \frac{1+i}{1+i} = \frac{(1+i)^{2}}{1^{2}-i^{2}} = \frac{1+i^{2}+2i}{1-(-1)} = \frac{1-1+2i}{2} = \frac{2i}{2} = i$.
Now,the equation becomes $i^{m} = 1$.
We know that the powers of $i$ follow a cycle:
$i^{1} = i$
$i^{2} = -1$
$i^{3} = -i$
$i^{4} = 1$
Thus,the least positive integral value of $m$ for which $i^{m} = 1$ is $m = 4$.

(A) Given $\left(\frac{1-i}{1+i}\right)^k = -i$.
First,simplify the base: $\frac{1-i}{1+i} = \frac{(1-i)(1-i)}{(1+i)(1-i)} = \frac{1-2i+i^2}{1-i^2} = \frac{1-2i-1}{1+1} = \frac{-2i}{2} = -i$.
So,the equation becomes $(-i)^k = -i$.
For $(-i)^k = -i$,$k$ must satisfy $k \equiv 1 \pmod 4$.
The least positive integer $m$ is $1$.
The greatest negative integer $n$ is $1 - 4 = -3$.
Thus,$m - n = 1 - (-3) = 1 + 3 = 4$.

(D) Given the sum $\sum_{k=0}^{40} i^k = x + iy$.
Since the sum of four consecutive powers of $i$ is zero,i.e.,$i^n + i^{n+1} + i^{n+2} + i^{n+3} = 0$,we can group the terms.
The sum has $41$ terms from $k=0$ to $k=40$.
$\sum_{k=0}^{40} i^k = i^0 + (i^1 + i^2 + i^3 + i^4) + \dots + (i^{37} + i^{38} + i^{39} + i^{40}) = 1 + 0 + \dots + 0 = 1$.
Thus,$x + iy = 1 + 0i$,which gives $x = 1$ and $y = 0$.
Substituting these values into the expression:
$x^{100} + x^{99}y + x^{242}y^2 + x^{97}y^3 = (1)^{100} + (1)^{99}(0) + (1)^{242}(0)^2 + (1)^{97}(0)^3 = 1 + 0 + 0 + 0 = 1$.

(C) Given expression: $i^{18}-3i^7+i^2(1+i^4)(i)^{22}$
We know that $i^2 = -1$,$i^3 = -i$,$i^4 = 1$.
$i^{18} = (i^4)^4 \times i^2 = (1)^4 \times (-1) = -1$
$i^7 = (i^4) \times i^3 = 1 \times (-i) = -i$
$i^{22} = (i^4)^5 \times i^2 = (1)^5 \times (-1) = -1$
Substituting these values:
$-1 - 3(-i) + (-1)(1+1)(-1)$
$= -1 + 3i + (-1)(2)(-1)$
$= -1 + 3i + 2$
$= 1 + 3i$

(B) Given $(x+iy) = \left(\frac{1+i}{1-i}\right)^3 - \left(\frac{1-i}{1+i}\right)^3$.
First,simplify the base fractions:
$\frac{1+i}{1-i} = \frac{(1+i)(1+i)}{(1-i)(1+i)} = \frac{1+2i+i^2}{1-i^2} = \frac{1+2i-1}{1+1} = \frac{2i}{2} = i$.
Similarly,$\frac{1-i}{1+i} = \frac{1}{i} = -i$.
Substituting these into the expression:
$x+iy = (i)^3 - (-i)^3$.
$x+iy = -i - (-i^3) = -i - (i) = -2i$.
Comparing the real and imaginary parts:
$x = 0$ and $y = -2$.
Since $0 > -2$,we have $x > y$.

(A) Given that,$\left(\frac{1+i}{1-i}\right)^m=1$.
First,simplify the base:
$\frac{1+i}{1-i} = \frac{1+i}{1-i} \times \frac{1+i}{1+i} = \frac{(1+i)^2}{1^2 - i^2} = \frac{1+i^2+2i}{1-(-1)} = \frac{1-1+2i}{2} = \frac{2i}{2} = i$.
So,the equation becomes $i^m = 1$.
We know that $i^n = 1$ if and only if $n$ is a multiple of $4$.
Checking the options:
$1934 \div 4 = 483.5$ (not a multiple of $4$).
$2024 \div 4 = 506$ (multiple of $4$).
$2172 \div 4 = 543$ (multiple of $4$).
$10^{100} = (2 \times 5)^{100} = 2^{100} \times 5^{100}$,which is divisible by $4$ since $2^{100}$ is divisible by $4$.
Thus,$m$ cannot be $1934$.

(B) Given the expression $\left(\frac{1+i}{1-i}\right)^n=1$.
First,simplify the base $\frac{1+i}{1-i}$ by multiplying the numerator and denominator by the conjugate $(1+i)$:
$\frac{1+i}{1-i} \times \frac{1+i}{1+i} = \frac{1+i^2+2i}{1-i^2} = \frac{1-1+2i}{1-(-1)} = \frac{2i}{2} = i$.
Thus,the equation becomes $i^n = 1$.
We know that $i^n = 1$ if and only if $n$ is a multiple of $4$.
We need to find the number of multiples of $4$ in the range $1 \leq n \leq 2021$.
The multiples are $4, 8, 12, \ldots, 2020$.
This is an arithmetic progression where $a = 4$,$d = 4$,and $l = 2020$.
Using the formula $l = a + (n-1)d$:
$2020 = 4 + (n-1)4$ $\Rightarrow 2016 = (n-1)4$ $\Rightarrow n-1 = 504$ $\Rightarrow n = 505$.
Therefore,there are $505$ such natural numbers.

(B) We know that the sum of four consecutive powers of $i$ is $i^k + i^{k+1} + i^{k+2} + i^{k+3} = i^k(1 + i - 1 - i) = 0$.
For the first sum: $\sum_{n=2}^{30} i^n = i^2 + i^3 + \dots + i^{30}$. The number of terms is $30 - 2 + 1 = 29$.
Since $29 = 4 \times 7 + 1$,the sum is $i^2 + (i^3 + i^4 + i^5 + i^6) + \dots + (i^{27} + i^{28} + i^{29} + i^{30}) = -1 + 0 = -1$.
For the second sum: $\sum_{n=30}^{65} i^{n+3} = i^{33} + i^{34} + \dots + i^{68}$. The number of terms is $68 - 33 + 1 = 36$.
Since $36$ is a multiple of $4$,the sum of these $36$ consecutive powers of $i$ is $0$.
Thus,the total sum is $-1 + 0 = -1$.

(D) Let $Z = \frac{(1-i)^3}{(2-i)(3-2i)}$.
First,expand the numerator: $(1-i)^3 = 1^3 - 3(1)^2(i) + 3(1)(i)^2 - i^3 = 1 - 3i - 3 + i = -2 - 2i$.
Next,expand the denominator: $(2-i)(3-2i) = 6 - 4i - 3i + 2i^2 = 6 - 7i - 2 = 4 - 7i$.
So,$Z = \frac{-2 - 2i}{4 - 7i}$.
To simplify,multiply the numerator and denominator by the conjugate of the denominator: $Z = \frac{-2 - 2i}{4 - 7i} \times \frac{4 + 7i}{4 + 7i} = \frac{-8 - 14i - 8i - 14i^2}{16 + 49} = \frac{-8 - 22i + 14}{65} = \frac{6 - 22i}{65}$.
Thus,$Z = \frac{6}{65} - \frac{22}{65}i$.
The imaginary part is $-\frac{22}{65}$.

(A) Given $z = (1-i)^3(x+i)$.
First,expand $(1-i)^3$:
$(1-i)^3 = 1^3 - 3(1)^2(i) + 3(1)(i)^2 - i^3 = 1 - 3i - 3 + i = -2 - 2i$.
Now,substitute this into the expression for $z$:
$z = (-2 - 2i)(x + i) = -2x - 2i - 2ix - 2i^2 = -2x - 2i - 2ix + 2 = (2 - 2x) - i(2 + 2x)$.
For $z$ to be purely imaginary,the real part must be zero:
$2 - 2x_1 = 0 \Rightarrow x_1 = 1$.
For $z$ to be purely real,the imaginary part must be zero:
$-(2 + 2x_2) = 0 \Rightarrow x_2 = -1$.
Therefore,$x_1 x_2 = 1 \times (-1) = -1$.

(A) Given: $\frac{x-1}{3+i} + \frac{y-1}{3-i} = i$
Multiply by the common denominator $(3+i)(3-i) = 3^2 + 1^2 = 10$:
$(x-1)(3-i) + (y-1)(3+i) = i(10)$
$3x - ix - 3 + i + 3y + iy - 3 - i = 10i$
$(3x + 3y - 6) + i(y - x) = 10i$
Comparing real and imaginary parts:
$3x + 3y - 6 = 0 \Rightarrow x + y = 2$
$y - x = 10$
Adding the two equations: $2y = 12 \Rightarrow y = 6$
Substituting $y = 6$ into $x + y = 2$: $x + 6 = 2 \Rightarrow x = -4$
Thus,$x = -4$ and $y = 6$.

(A) Given: $\frac{(1+i)x-2i}{3+i} + \frac{(2-3i)y}{3-i} = i$
Multiply by $(3+i)(3-i) = 3^2 - i^2 = 9 - (-1) = 10$:
$((1+i)x-2i)(3-i) + (2-3i)y(3+i) = 10i$
$(3x - ix + 3ix - i^2x - 6i + 2i^2) + y(6 + 2i - 9i - 3i^2) = 10i$
$(3x - ix + 3ix + x - 6i - 2) + y(6 - 7i + 3) = 10i$
$(4x + 2ix - 2 - 6i) + y(9 - 7i) = 10i$
$(4x - 2 + 9y) + i(2x - 6 - 7y) = 10i$
Equating real and imaginary parts:
Real part: $4x + 9y - 2 = 0 \Rightarrow 4x + 9y = 2$
Imaginary part: $2x - 7y - 6 = 10 \Rightarrow 2x - 7y = 16$
Solving the system:
Multiply the second equation by $2$: $4x - 14y = 32$
Subtract from the first: $(4x + 9y) - (4x - 14y) = 2 - 32$
$23y = -30 \Rightarrow y = -30/23$
Substitute $y$ into $2x - 7y = 16$:
$2x - 7(-30/23) = 16 \Rightarrow 2x + 210/23 = 368/23$
$2x = 158/23 \Rightarrow x = 79/23$
Therefore,$x+y = 79/23 - 30/23 = 49/23$.

(A) We know that $i^2 = -1$,$i^4 = 1$.
First,calculate $i^{22}$: $i^{22} = (i^4)^5 \times i^2 = 1^5 \times (-1) = -1$.
Next,calculate $(\frac{1}{i})^{35}$: $\frac{1}{i} = -i$.
So,$(-i)^{35} = -(i^{35}) = -(i^{32} \times i^3) = -(1 \times -i) = i$.
Now,substitute these values into the expression: $\{i^{22} - (\frac{1}{i})^{35}\}^2 = \{-1 - i\}^2$.
Expand the square: $(-1 - i)^2 = (-1)^2 + (-i)^2 + 2(-1)(-i) = 1 + i^2 + 2i$.
Since $i^2 = -1$,we get $1 - 1 + 2i = 2i$.

(D) Given,$a+bi = \frac{i}{1-i}$
Multiply the numerator and denominator by the conjugate of the denominator $(1+i)$:
$a+bi = \frac{i(1+i)}{(1-i)(1+i)}$
$a+bi = \frac{i+i^2}{1^2-i^2}$
Since $i^2 = -1$:
$a+bi = \frac{i-1}{1-(-1)} = \frac{-1+i}{2}$
$a+bi = \frac{-1}{2} + \frac{1}{2}i$
Comparing the real and imaginary parts,we get $a = \frac{-1}{2}$ and $b = \frac{1}{2}$
Therefore,$(a, b) = (\frac{-1}{2}, \frac{1}{2})$

(B) To find the quadrant,we first simplify the complex number by multiplying the numerator and denominator by the conjugate of the denominator,which is $1+i$:
$\frac{1+2i}{1-i} \times \frac{1+i}{1+i} = \frac{1+i+2i+2i^2}{1^2-i^2}$
Since $i^2 = -1$,we have:
$\frac{1+3i-2}{1-(-1)} = \frac{-1+3i}{2} = -\frac{1}{2} + \frac{3}{2}i$
The real part is $-\frac{1}{2}$ (negative) and the imaginary part is $\frac{3}{2}$ (positive).
$A$ complex number with a negative real part and a positive imaginary part lies in the Second quadrant.

(D) Given $(x+iy)^{\frac{1}{3}} = 5+3i$.
Cubing both sides,we get $x+iy = (5+3i)^3$.
Using the identity $(a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3$,we have:
$x+iy = 5^3 + 3(5^2)(3i) + 3(5)(3i)^2 + (3i)^3$.
$x+iy = 125 + 3(25)(3i) + 15(9i^2) + 27i^3$.
Since $i^2 = -1$ and $i^3 = -i$,we get:
$x+iy = 125 + 225i + 135(-1) + 27(-i)$.
$x+iy = 125 + 225i - 135 - 27i$.
$x+iy = (125-135) + (225-27)i$.
$x+iy = -10 + 198i$.
Comparing real and imaginary parts,$x = -10$ and $y = 198$.
Now,calculate $3x+5y = 3(-10) + 5(198) = -30 + 990 = 960$.

(B) First,simplify each term individually:
$1$. $\frac{5+2i}{2-5i} = \frac{(5+2i)(2+5i)}{(2-5i)(2+5i)} = \frac{10 + 25i + 4i + 10i^2}{4 + 25} = \frac{10 + 29i - 10}{29} = \frac{29i}{29} = i$
$2$. $\frac{3-4i}{4+3i} = \frac{(3-4i)(4-3i)}{(4+3i)(4-3i)} = \frac{12 - 9i - 16i + 12i^2}{16 + 9} = \frac{12 - 25i - 12}{25} = \frac{-25i}{25} = -i$
$3$. $\frac{1}{i} = \frac{1 \times i}{i \times i} = \frac{i}{-1} = -i$
Substituting these back into the expression for $z$:
$z = (i) - (-i) - (-i) = i + i + i = 3i$
The complex number is $z = 0 + 3i$.
Thus,the real part of $z$ is $0$.

(C) First,simplify the term $\frac{1+i}{1-i}$ by multiplying the numerator and denominator by the conjugate $(1+i)$:
$\frac{1+i}{1-i} = \frac{(1+i)(1+i)}{(1-i)(1+i)} = \frac{1+i^2+2i}{1-i^2} = \frac{1-1+2i}{1+1} = \frac{2i}{2} = i$
Similarly,for the second term:
$\frac{1-i}{1+i} = \frac{(1-i)(1-i)}{(1+i)(1-i)} = \frac{1+i^2-2i}{1-i^2} = \frac{1-1-2i}{1+1} = \frac{-2i}{2} = -i$
Now,substitute these values into the original expression:
$(i)^4 + (-i)^4 = i^4 + i^4 = 1 + 1 = 2$

(A) Given the equation: $\frac{(1+i) x-i}{2+i}+\frac{(1+2 i) y+i}{2-i}=1$
Multiply the first term by $\frac{2-i}{2-i}$ and the second term by $\frac{2+i}{2+i}$:
$\frac{[(1+i)x-i](2-i)}{5} + \frac{[(1+2i)y+i](2+i)}{5} = 1$
Expanding the numerators:
$\frac{(2-i+2i-i^2)x - 2i + i^2}{5} + \frac{(2+i+4i+2i^2)y + 2i + i^2}{5} = 1$
Since $i^2 = -1$:
$\frac{(3+i)x - 2i - 1}{5} + \frac{(5i)y + 2i - 1}{5} = 1$
$(3+i)x + (5i)y - 2 = 5$
$(3x-7) + i(x+5y) = 0$
Comparing real and imaginary parts:
$3x-7 = 0 \Rightarrow x = \frac{7}{3}$
$x+5y = 0 \Rightarrow y = -\frac{x}{5} = -\frac{7}{15}$
Thus,$(x, y) = \left(\frac{7}{3}, -\frac{7}{15}\right)$.

(B) Given that $-3+ix^2y$ and $x^2+y+4i$ are complex conjugates.
Therefore,$-3-ix^2y = x^2+y+4i$.
Comparing the real and imaginary parts on both sides:
$x^2+y = -3$ $(i)$
$-x^2y = 4$ $(ii)$
From $(ii)$,$y = -\frac{4}{x^2}$.
Substituting $y$ in $(i)$:
$x^2 - \frac{4}{x^2} = -3$
Let $x^2 = t$,then $t - \frac{4}{t} = -3 \implies t^2 + 3t - 4 = 0$.
$(t+4)(t-1) = 0$.
Since $t = x^2$ must be positive,$t = 1$.
$x^2 = 1 \implies x = \pm 1$.

(B) Given,$u+iv = \frac{3i}{(x+2)+iy}$.
Taking the reciprocal on both sides:
$\frac{1}{u+iv} = \frac{(x+2)+iy}{3i}$.
Multiply the numerator and denominator of the left side by the conjugate $(u-iv)$:
$\frac{u-iv}{u^2+v^2} = \frac{(x+2)+iy}{3i}$.
Multiply both sides by $3i$:
$(x+2)+iy = \frac{3i(u-iv)}{u^2+v^2} = \frac{3ui - 3vi^2}{u^2+v^2} = \frac{3v + 3ui}{u^2+v^2}$.
Equating the imaginary parts:
$y = \frac{3u}{u^2+v^2}$.

(B) We know that $(1+i)^2 = 1+i^2+2i = 1-1+2i = 2i$ and $(1-i)^2 = 1+i^2-2i = 1-1-2i = -2i$.
Substituting these into the expression:
$(1+i)^{2024} + (1-i)^{2024} = [(1+i)^2]^{1012} + [(1-i)^2]^{1012}$
$= (2i)^{1012} + (-2i)^{1012}$
$= 2^{1012} \cdot i^{1012} + (-2)^{1012} \cdot i^{1012}$
Since $1012$ is an even number,$(-2)^{1012} = 2^{1012}$ and $i^{1012} = (i^4)^{253} = 1^{253} = 1$.
$= 2^{1012} \cdot 1 + 2^{1012} \cdot 1$
$= 2 \cdot 2^{1012} = 2^{1013}$.

(D) Let $z = \frac{3+2 i \sin \theta}{1-2 i \sin \theta}$.
To make $z$ real,the imaginary part of $z$ must be zero.
Multiply the numerator and denominator by the conjugate of the denominator $(1+2 i \sin \theta)$:
$z = \frac{(3+2 i \sin \theta)(1+2 i \sin \theta)}{(1-2 i \sin \theta)(1+2 i \sin \theta)}$
$z = \frac{3 + 6 i \sin \theta + 2 i \sin \theta + 4 i^2 \sin^2 \theta}{1 + 4 \sin^2 \theta}$
Since $i^2 = -1$:
$z = \frac{(3 - 4 \sin^2 \theta) + i(8 \sin \theta)}{1 + 4 \sin^2 \theta}$
For $z$ to be real,the imaginary part must be zero:
$\frac{8 \sin \theta}{1 + 4 \sin^2 \theta} = 0$
$8 \sin \theta = 0$
$\sin \theta = 0$
Therefore,$\theta = n \pi$ for $n \in \mathbb{Z}$.

(D) First,simplify the base expressions:
$\frac{1-i}{1+i} = \frac{(1-i)(1-i)}{(1+i)(1-i)} = \frac{1-2i+i^2}{1-i^2} = \frac{1-2i-1}{1+1} = \frac{-2i}{2} = -i$
$\frac{1+i}{1-i} = \frac{(1+i)(1+i)}{(1-i)(1+i)} = \frac{1+2i+i^2}{1-i^2} = \frac{1+2i-1}{1+1} = \frac{2i}{2} = i$
Now substitute these into the expression:
$(-i)^{2022} + (i)^{2021}$
$= (i)^{2022} + (i)^{2021}$
$= (i^{4})^{505} \cdot i^2 + (i^{4})^{505} \cdot i^1$
$= (1)^{505} \cdot (-1) + (1)^{505} \cdot i$
$= -1 + i$

(C) The given series is $S = 1+i^2+i^4+i^6+\ldots+i^{2024}$.
We know that $i^2 = -1$,$i^4 = 1$,$i^6 = -1$,and so on.
This is a geometric progression with first term $a=1$,common ratio $r=i^2=-1$,and the number of terms $n = \frac{2024-0}{2} + 1 = 1013$.
The sum of a geometric series is $S_n = \frac{a(r^n - 1)}{r - 1}$.
Substituting the values: $S = \frac{1((-1)^{1013} - 1)}{-1 - 1} = \frac{-1 - 1}{-2} = \frac{-2}{-2} = 1$.

(B) Given,$\frac{(1+i)^n}{(1-i)^n} = -i$
$\Rightarrow \left(\frac{1+i}{1-i}\right)^n = -i$
Rationalizing the denominator inside the bracket:
$\Rightarrow \left[\frac{(1+i)(1+i)}{(1-i)(1+i)}\right]^n = -i$
$\Rightarrow \left[\frac{1+i^2+2i}{1-i^2}\right]^n = -i$
Since $i^2 = -1$:
$\Rightarrow \left[\frac{1-1+2i}{1+1}\right]^n = -i$
$\Rightarrow \left(\frac{2i}{2}\right)^n = -i$
$\Rightarrow i^n = -i$
We know that $i^1 = i$,$i^2 = -1$,$i^3 = -i$,and $i^4 = 1$.
The value $i^n = -i$ occurs when $n$ is of the form $4k-1$ for $k \in N$ (e.g.,for $k=1, n=3$; for $k=2, n=7$).

(D) We know that the sum of four consecutive powers of $i$ is zero,i.e.,$i^n+i^{n+1}+i^{n+2}+i^{n+3}=0$ for any integer $n$.
The given series is $S = i^2+i^3+i^4+\ldots+i^{4000}$.
This series contains $4000-2+1 = 3999$ terms.
We can group these terms into sets of four. Since $3999 = 4 \times 999 + 3$,there are $999$ groups of four consecutive powers of $i$ (each summing to $0$) and $3$ remaining terms.
The sum of the first $3996$ terms is $0$.
Thus,$S = i^{3997} + i^{3998} + i^{3999} + i^{4000} = 0$ is not the correct approach here; rather,we observe the sequence $i^2, i^3, i^4, \ldots, i^{4000}$.
The sum is $\sum_{k=2}^{4000} i^k = \frac{i^2(1-i^{3999})}{1-i} = \frac{-1(1-i^3)}{1-i} = \frac{-1(1-(-i))}{1-i} = \frac{-(1+i)}{1-i} = \frac{-(1+i)(1+i)}{(1-i)(1+i)} = \frac{-(1+2i+i^2)}{1-i^2} = \frac{-(1+2i-1)}{1+1} = \frac{-2i}{2} = -i$.

(D) We know that $(1+i)^2 = 1 + i^2 + 2i = 2i$ and $(1-i)^2 = 1 + i^2 - 2i = -2i$.
Given expression: $E = \frac{(1+i)^{2011}}{(1-i)^{2009}} = \frac{(1+i)^{2009} \cdot (1+i)^2}{(1-i)^{2009}}$.
$E = \left(\frac{1+i}{1-i}\right)^{2009} \cdot (1+i)^2$.
Simplify $\frac{1+i}{1-i}$: $\frac{1+i}{1-i} \cdot \frac{1+i}{1+i} = \frac{1+i^2+2i}{1-i^2} = \frac{2i}{2} = i$.
Substitute back: $E = (i)^{2009} \cdot (2i) = 2 \cdot i^{2010}$.
Since $i^4 = 1$,$i^{2010} = (i^4)^{502} \cdot i^2 = 1^{502} \cdot (-1) = -1$.
Therefore,$E = 2 \cdot (-1) = -2$.

(B) Given the expression $\frac{2+3 i}{i-2}-\frac{4 i-3}{3+4 i}=x+i y$.
First,simplify the first term: $\frac{2+3 i}{-2+i} \times \frac{-2-i}{-2-i} = \frac{-4-2i-6i-3i^2}{4+1} = \frac{-4-8i+3}{5} = \frac{-1-8i}{5} = -0.2 - 1.6i$.
Next,simplify the second term: $\frac{-3+4i}{3+4i} \times \frac{3-4i}{3-4i} = \frac{-9+12i+12i-16i^2}{9+16} = \frac{-9+24i+16}{25} = \frac{7+24i}{25} = 0.28 + 0.96i$.
Subtracting the two: $(-0.2 - 1.6i) - (0.28 + 0.96i) = -0.48 - 2.56i$.
Thus,$x = -0.48$ and $y = -2.56$.
Calculating $3x+y = 3(-0.48) + (-2.56) = -1.44 - 2.56 = -4$.

(C) 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,substitute this into the original expression:
$i^{228} = (i^4)^{57} = 1^{57} = 1$.
Next,evaluate the options to see which one equals $1$:
For option $C$: $\left(\frac{1-i}{1+i}\right)^{228} = \left(\frac{1}{i}\right)^{228} = \frac{1}{i^{228}} = \frac{1}{1} = 1$.
Thus,the correct option is $C$.

(B) Given $(a+ib)^{\frac{1}{4}}=2+3i$.
Raising both sides to the power of $4$,we get:
$(a+ib)=(2+3i)^4$.
Expanding the right side:
$(a+ib)=[(2+3i)^2]^2 = [4+9i^2+12i]^2$.
Since $i^2=-1$,we have:
$(a+ib)=[4-9+12i]^2 = [-5+12i]^2$.
Expanding further:
$a+ib = (-5)^2 + (12i)^2 + 2(-5)(12i) = 25 - 144 - 120i = -119 - 120i$.
Comparing the real and imaginary parts,we get $a=-119$ and $b=-120$.
Now,calculating $3b-2a$:
$3b-2a = 3(-120) - 2(-119) = -360 + 238 = -122$.

(B) Given,$\sqrt{x^2 - 2x + 8} + (x + 4)i = y(2 + i)$.
Equating real and imaginary parts:
$\sqrt{x^2 - 2x + 8} = 2y$ $(1)$
$x + 4 = y$ $(2)$
Substitute $y = x + 4$ into equation $(1)$:
$\sqrt{x^2 - 2x + 8} = 2(x + 4)$
Squaring both sides:
$x^2 - 2x + 8 = 4(x^2 + 8x + 16)$
$x^2 - 2x + 8 = 4x^2 + 32x + 64$
$3x^2 + 34x + 56 = 0$
$(3x + 28)(x + 2) = 0$
So,$x = -2$ or $x = -\frac{28}{3}$.
If $x = -2$,then $y = -2 + 4 = 2$. Thus,$Z = -2 + 2i$.
If $x = -\frac{28}{3}$,then $y = -\frac{28}{3} + 4 = -\frac{16}{3}$. Thus,$Z = -\frac{28}{3} - \frac{16}{3}i$.
Comparing with the given options,$Z = -2 + 2i$ is the correct choice.

(C) Given,$(1+i)^n=(1-i)^n$
$\Rightarrow \frac{(1+i)^n}{(1-i)^n}=1$
$\Rightarrow \left[\frac{(1+i)(1+i)}{(1-i)(1+i)}\right]^n=1$
$\Rightarrow \left[\frac{1+i^2+2i}{1-i^2}\right]^n=1$
$\Rightarrow \left[\frac{1-1+2i}{1+1}\right]^n=1$
$\Rightarrow \left(\frac{2i}{2}\right)^n=1$
$\Rightarrow i^n=1$
Since the smallest positive integer $n$ for which $i^n=1$ is $4$,therefore $n=4$.

(A) Let $z = \frac{3+2i \sin \theta}{1-2i \sin \theta}$. To simplify,multiply the numerator and denominator by the conjugate of the denominator $(1+2i \sin \theta)$:
$z = \frac{(3+2i \sin \theta)(1+2i \sin \theta)}{(1-2i \sin \theta)(1+2i \sin \theta)}$
$z = \frac{3 + 6i \sin \theta + 2i \sin \theta + 4i^2 \sin^2 \theta}{1^2 + (2 \sin \theta)^2}$
Since $i^2 = -1$,we have:
$z = \frac{3 - 4 \sin^2 \theta + 8i \sin \theta}{1 + 4 \sin^2 \theta}$
$z = \frac{3 - 4 \sin^2 \theta}{1 + 4 \sin^2 \theta} + i \left( \frac{8 \sin \theta}{1 + 4 \sin^2 \theta} \right)$
For $z$ to be a real number,the imaginary part must be zero:
$\frac{8 \sin \theta}{1 + 4 \sin^2 \theta} = 0$
This implies $\sin \theta = 0$.
Given $0 < \theta < 2\pi$,the only solution is $\theta = \pi$.

(D) Given $(2x - y + 1) + i(x - 2y - 1) = 2 - 3i$.
Comparing the real and imaginary parts,we get:
$2x - y + 1 = 2 \implies 2x - y = 1$ (Equation $1$)
$x - 2y - 1 = -3 \implies x - 2y = -2$ (Equation $2$)
Multiplying Equation $2$ by $2$,we get $2x - 4y = -4$ (Equation $3$).
Subtracting Equation $3$ from Equation $1$:
$(2x - y) - (2x - 4y) = 1 - (-4)
3y = 5 \implies y = \frac{5}{3}$.
Substituting $y = \frac{5}{3}$ in Equation $1$:
$2x - \frac{5}{3} = 1
2x = 1 + \frac{5}{3} = \frac{8}{3}
x = \frac{4}{3}$.
We need the multiplicative inverse of $(x - iy) = (\frac{4}{3} - i\frac{5}{3})$.
The inverse is $\frac{1}{\frac{4}{3} - i\frac{5}{3}} = \frac{3}{4 - 5i}$.
Rationalizing the denominator:
$\frac{3(4 + 5i)}{(4 - 5i)(4 + 5i)} = \frac{12 + 15i}{16 + 25} = \frac{12 + 15i}{41} = \frac{12}{41} + \frac{15}{41}i$.

(C) We have $(1+i)^2 = 1+i^2+2i = 1-1+2i = 2i$.
Similarly,$(1-i)^2 = 1+i^2-2i = 1-1-2i = -2i$.
Now,$(1+i)^{10} = ((1+i)^2)^5 = (2i)^5 = 2^5 \times i^5 = 32i$.
And $(1-i)^{10} = ((1-i)^2)^5 = (-2i)^5 = (-2)^5 \times i^5 = -32i$.
Therefore,$(1+i)^{10} + (1-i)^{10} = 32i + (-32i) = 0$.

(C) Given the determinant: $\left|\begin{array}{cc}1-i & i \\ 1+2 i & -i\end{array}\right|=x+i y$
Expanding the determinant: $(1-i)(-i) - (i)(1+2i) = x+iy$
$-i + i^2 - (i + 2i^2) = x+iy$
Since $i^2 = -1$,we substitute: $-i - 1 - (i - 2) = x+iy$
$-i - 1 - i + 2 = x+iy$
$1 - 2i = x+iy$
Comparing the real and imaginary parts,we get $x = 1$ and $y = -2$.

(B) We have $\sum_{n=1}^{13}(i^{n}+i^{n+1}) = \sum_{n=1}^{13} i^{n} + \sum_{n=1}^{13} i^{n+1}$.
Since $i^{n}$ is a geometric progression with first term $a=i$ and common ratio $r=i$,the sum of $13$ terms is $S_{13} = i \frac{1-i^{13}}{1-i}$.
Note that $i^{13} = (i^{4})^{3} \times i = 1^{3} \times i = i$.
Thus,$\sum_{n=1}^{13} i^{n} = i \frac{1-i}{1-i} = i$.
Similarly,$\sum_{n=1}^{13} i^{n+1} = i^{2} \frac{1-i^{13}}{1-i} = -1 \frac{1-i}{1-i} = -1$.
Therefore,the total sum is $i + (-1) = i-1$.