Imaginary Numbers: Practice

Divide the exponent by $4$ and keep the remainder: $2023 = 4 \times 505 + 3$.

$i^{2023} = i^3 = -i$

Convert first, then square: $\sqrt{-6} = i\sqrt{6}$.

$\sqrt{-6}\cdot\sqrt{-6} = (i\sqrt{6})^2 = 6 i^2 = -6$

Multiplying the radicands to get $\sqrt{36} = 6$ is the trap; both radicands are negative, so the true value is $-6$.

Reduce each exponent modulo $4$. The powers $i^5$, $i^9$, and $i^{13}$ all leave remainder $1$, so each equals $i$. But $i^3$ leaves remainder $3$.

$i^3 = -i \ne i$

So $i^3$ is the one that is not equal to $i$.

Raise the $2$ and the $i$ to the fourth power separately, then use $i^4 = 1$.

$(2i)^4 = 2^4 \cdot i^4 = 16 \cdot 1 = 16$

Replace each power with its value: $i^4 = 1$ and $i^2 = -1$.

$i^4 - i^2 = 1 - (-1) = 2$

Add the exponents first: $i^{50}\cdot i^{51} = i^{101}$. Then $101 = 4 \times 25 + 1$.

$i^{50}\cdot i^{51} = i^{101} = i^1 = i$

Write out the four consecutive powers and add.

$i + (-1) + (-i) + 1 = 0$

The $i$ and $-i$ cancel, and the $-1$ and $1$ cancel, so any four powers in a row sum to $0$.

Isolate the square, then take the root and simplify with $12 = 4 \times 3$.

$x^2 = -12 \quad\Rightarrow\quad x = \pm\sqrt{-12} = \pm 2i\sqrt{3}$

Divide by $4$ and keep the remainder: $99 = 4 \times 24 + 3$.

$i^{99} = i^3 = -i$

Rewrite each root with $i$ before multiplying: $\sqrt{-16} = 4i$ and $\sqrt{-25} = 5i$.

$\sqrt{-16}\cdot\sqrt{-25} = (4i)(5i) = 20 i^2 = -20$

Not $\sqrt{400} = 20$, because both radicands are negative.

An exponent that is a multiple of $4$ leaves remainder $0$.

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

Simplify each root first, then combine the like $i$ terms.

$\sqrt{-9} + \sqrt{-16} = 3i + 4i = 7i$

Roots do not add inside a single radical, so it is not $\sqrt{-25} = 5i$.