Q.Find the conjugate of $\dfrac{(3 - 2i)(2 + 3i)}{(1 + 2i)(2 - i)}$.

$(1 + i^2) = 0$, so the entire product equals $0$.

Recall $i^2 = -1$, $i^3 = -i$, $i^4 = 1$.

So the second factor is $1 + i^2 = 1 - 1 = 0$.

A product containing a zero factor is $0$, regardless of the other factors.

$(1+i)(0)(1+i^3)(1+i^4) = 0$.

True — complex number multiplication is commutative, just like real number multiplication.

Let $z_1 = a+ib$ and $z_2 = c+id$ be any two complex numbers. Then

$z_1 z_2 = (a+ib)(c+id) = (ac-bd) + i(ad+bc)$

$z_2 z_1 = (c+id)(a+ib) = (ca-db) + i(cb+da) = (ac-bd)+i(ad+bc)$

Since $ac-bd=ca-db$ and $ad+bc=cb+da$ (real number multiplication and addition are commutative), $z_1z_2=z_2z_1$ for all complex numbers. The commutative law of multiplication holds in $\mathbb{C}$ just as it does in $\mathbb{R}$.

$i^{4k}=1$ for every integer $k$.

Recall $i^1=i$, $i^2=-1$, $i^3=-i$, $i^4=1$, and then the pattern repeats every 4 powers since $i^4=1$. So for any integer $k$,

$i^{4k} = (i^4)^k = 1^k = 1$

This holds for negative $k$ too, since $i^{-4}=\dfrac{1}{i^4}=\dfrac{1}{1}=1$.

Powers of $i$ repeat every 4 steps, so reduce the exponent mod 4.

Recall $i^2 = -1$, $i^3=-i$, $i^4=1$, and then the cycle repeats. So $i^n$ depends only on $n \bmod 4$.

$2022 = 4 \times 505 + 2$, so $2022 \equiv 2 \pmod 4$.

Hence $i^{2022} = i^2 = -1$.

In $a+ib$ form this is $-1 + 0i$.

Powers of $i$ repeat with period $4$: $i^1=i,\ i^2=-1,\ i^3=-i,\ i^4=1$.

To evaluate $i^{-37}$, reduce the exponent modulo $4$. Since $-37 = 4(-10) + 3$, we have $-37 \equiv 3 \pmod 4$, so

$i^{-37} = i^{3} = -i$

(Check directly: $i^{-37} = \dfrac{1}{i^{37}}$, and $37 = 4(9)+1$ so $i^{37}=i^1=i$; thus $i^{-37} = \dfrac{1}{i} = \dfrac{1}{i}\cdot\dfrac{-i}{-i} = \dfrac{-i}{-i^2} = \dfrac{-i}{1} = -i$, matching.)

So $i^{-37} = 0 + (-1)i$, which has real part $0$.

The multiplicative inverse (reciprocal) of a nonzero complex number $z$ is $1/z$; for a purely imaginary unit like $-i$, rationalise by multiplying by its conjugate.

We need $(-i)^{-1} = \dfrac{1}{-i}$.

Multiply numerator and denominator by $i$ (rationalising, since $i \cdot (-i)$ conveniently clears the imaginary denominator):

$\frac{1}{-i} = \frac{1}{-i}\times\frac{i}{i} = \frac{i}{-i^2} = \frac{i}{-(-1)} = \frac{i}{1} = i$

Verify: $(-i)\times i = -i^2 = -(-1) = 1$ ✓, confirming $i$ is indeed the multiplicative inverse of $-i$.

Cube the binomial step by step; the result is $-10-198i$.

We compute $(5-3i)^3$ in two stages.

Step 1 — square first:

$(5-3i)^2 = 25 - 2(5)(3i) + (3i)^2 = 25 - 30i + 9i^2 = 25 - 30i - 9 = 16 - 30i$

(using $i^2=-1$).

Step 2 — multiply by $(5-3i)$ once more:

$(5-3i)(16-30i) = 5(16) + 5(-30i) - 3i(16) - 3i(-30i)$

$= 80 - 150i - 48i + 90i^2 = 80 - 198i - 90 = -10 - 198i$

So $(5-3i)^3 = -10 - 198i$, already in the form $x+iy$ with $x=-10$, $y=-198$.

The two complex (non-real) cube roots of $1$ are $\dfrac{-1+i\sqrt3}{2}$ and $\dfrac{-1-i\sqrt3}{2}$.

We solve $x^3=1 \implies x^3-1=0 \implies (x-1)(x^2+x+1)=0$.

One root is $x=1$ (real). The other two come from $x^2+x+1=0$:

$x=\dfrac{-1\pm\sqrt{1-4}}{2}=\dfrac{-1\pm i\sqrt3}{2}.$

These are conventionally called $\omega=\dfrac{-1+i\sqrt3}{2}$ and $\omega^2=\dfrac{-1-i\sqrt3}{2}$ (each is the square of the other, and $1+\omega+\omega^2=0$).