Imaginary Numbers: Practice

Reduce the exponent modulo $4$: $10 = 4 \times 2 + 2$, remainder $2$.

$i^{10} = i^2 = -1$

Reduce the exponent modulo $4$: $15 = 4 \times 3 + 3$, remainder $3$.

$i^{15} = i^3 = -i$

Factor out the $i$, then simplify the radical with $12 = 4 \times 3$.

$\sqrt{-12} = i\sqrt{12} = i\sqrt{4 \times 3} = 2i\sqrt{3}$

Factor out the $i$, then simplify the radical with $50 = 25 \times 2$.

$\sqrt{-50} = i\sqrt{50} = i\sqrt{25 \times 2} = 5i\sqrt{2}$

Rewrite each root using $i$ before multiplying, since the product rule fails for two negatives.

$\sqrt{-4}\cdot\sqrt{-9} = (2i)(3i) = 6i^2 = -6$

Multiplying the radicands to get $\sqrt{36} = 6$ is the trap and has the wrong sign.

Multiply the real factors, multiply the $i$'s, then use $i^2 = -1$.

$(5i)(2i) = 10 i^2 = -10$

Factor out the $i$, then simplify the radical with $18 = 9 \times 2$.

$\sqrt{-18} = i\sqrt{18} = i\sqrt{9 \times 2} = 3i\sqrt{2}$

Convert each root first: $\sqrt{-2} = i\sqrt{2}$ and $\sqrt{-8} = 2i\sqrt{2}$.

$\sqrt{-2}\cdot\sqrt{-8} = (i\sqrt{2})(2i\sqrt{2}) = 2i^2(\sqrt{2}\cdot\sqrt{2}) = 2(-1)(2) = -4$

The shortcut $\sqrt{16} = 4$ has the wrong sign because both radicands are negative.

Reduce the exponent modulo $4$: $13 = 4 \times 3 + 1$, remainder $1$.

$i^{13} = i^1 = i$

A square root of a negative is a real multiple of $i$.

$\sqrt{-5} = i\sqrt{5}$

The others are real: $\sqrt{5}$ is real, $i^2 = -1$, and $-4$ is real.

Write $\sqrt{-3} = i\sqrt{3}$ and square it, using $i^2 = -1$.

$\left(\sqrt{-3}\right)^2 = (i\sqrt{3})^2 = 3 i^2 = -3$

Directly, squaring a square root returns the radicand, which is $-3$.

Reduce the exponent modulo $4$: $7 = 4 + 3$, remainder $3$.

$i^7 = i^3 = -i$