Q.Express $\left(-\sqrt{3} + \sqrt{-2}\right)\left(2\sqrt{3} - i\right)$ in the form of $a + ib$.

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

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

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

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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

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

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

$i^{4k}=1$ 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 $k$,

$i^{4k} = (i^4)^k = 1^k = 1$ …

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 $2022 \equiv 2 \pmod 4$.

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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

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

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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):

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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:

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

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 $x^2+x+1=0$:

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