Imaginary Numbers: Additional Practice (Set 1)

Reduce the exponent modulo $4$: $38 = 4 \times 9 + 2$, remainder $2$.

$i^{38} = i^2 = -1$

Convert each root first: $\sqrt{-3} = i\sqrt{3}$ and $\sqrt{-12} = 2i\sqrt{3}$.

$\sqrt{-3}\cdot\sqrt{-12} = (i\sqrt{3})(2i\sqrt{3}) = 2i^2(3) = -6$

The shortcut $\sqrt{36} = 6$ has the wrong sign, since both radicands are negative.

Reduce the exponent modulo $4$: $45 = 4 \times 11 + 1$, remainder $1$.

$i^{45} = i^1 = i$

Isolate the square, then take the root and simplify with $8 = 4 \times 2$.

$x^2 = -8 \quad\Rightarrow\quad x = \pm\sqrt{-8} = \pm 2i\sqrt{2}$

Square the $4$ and the $i$ separately, then use $i^2 = -1$.

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

Evaluate each. Now $i^2 = -1$, which is real, not imaginary.

$i^2 = -1$

The others are real multiples of $i$: $3i$, $-i$, and $\sqrt{-4} = 2i$.

Add the exponents first, then reduce: $i^4 \cdot i^6 = i^{10}$, and $10$ leaves remainder $2$.

$i^4 \cdot i^6 = i^{10} = i^2 = -1$

Reduce the exponent modulo $4$: $31 = 4 \times 7 + 3$, remainder $3$.

$i^{31} = i^3 = -i$

The exponent $200$ is a multiple of $4$, so the remainder is $0$.

$i^{200} = (i^4)^{50} = 1$

Take the square root of both sides, keeping both signs.

$x = \pm\sqrt{-100} = \pm 10i$

Rewrite each root using $i$ before multiplying: $\sqrt{-4} = 2i$ and $\sqrt{-9} = 3i$.

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

The trap answer $\sqrt{36} = 6$ ignores that both radicands are negative.

Evaluate each square first: $(2i)^2 = -4$ and $(3i)^2 = -9$.

$(2i)^2 (3i)^2 = (-4)(-9) = 36$

The two negatives multiply to a positive.