Imaginary Numbers: Additional Practice (Set 2)

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

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

Factor out the $i$, then simplify the radical with $98 = 49 \times 2$.

$\sqrt{-98} = i\sqrt{98} = i\sqrt{49 \times 2} = 7i\sqrt{2}$

Factor out the $i$, then simplify the radical with $45 = 9 \times 5$.

$\sqrt{-45} = i\sqrt{45} = i\sqrt{9 \times 5} = 3i\sqrt{5}$

Simplify the root and use $i^2 = -1$: $\sqrt{-9} = 3i$.

$i^2 \cdot \sqrt{-9} = (-1)(3i) = -3i$

Cube the $3$ and the $i$ separately, using $i^3 = -i$.

$(3i)^3 = 3^3 \cdot i^3 = 27(-i) = -27i$

Convert first, then square: $\sqrt{-5} = i\sqrt{5}$.

$\sqrt{-5}\cdot\sqrt{-5} = (i\sqrt{5})^2 = 5 i^2 = -5$

Not $\sqrt{25} = 5$, because both radicands are negative.

Reduce each exponent modulo $4$. The exponents $4$, $8$, and $12$ are all multiples of $4$, so those powers equal $1$. But $6$ leaves remainder $2$.

$i^6 = i^2 = -1$

Factor out the $i$, then simplify the radical with $27 = 9 \times 3$.

$\sqrt{-27} = i\sqrt{27} = i\sqrt{9 \times 3} = 3i\sqrt{3}$

Square the $i$ and the $\sqrt{5}$ separately, using $i^2 = -1$.

$\left(i\sqrt{5}\right)^2 = i^2 \left(\sqrt{5}\right)^2 = (-1)(5) = -5$

Reduce the exponent modulo $4$: $21 = 4 \times 5 + 1$, remainder $1$.

$i^{21} = i^1 = i$

Each exponent is a multiple of $4$, so each power equals $1$.

$i^4 + i^8 + i^{12} = 1 + 1 + 1 = 3$

Isolate the square, then take the root and simplify with $45 = 9 \times 5$.

$x^2 = -45 \quad\Rightarrow\quad x = \pm\sqrt{-45} = \pm 3i\sqrt{5}$