Imaginary Numbers: Practice

The imaginary unit $i$ is defined as the number whose square is $-1$.

$i^2 = -1$

That single equation is the entire definition of $i$.

The imaginary unit is built so that its square is $-1$, filling the gap that no real number can, since a real square is never negative.

$i^2 = -1$

Equivalently, $i = \sqrt{-1}$.

By definition, $i$ is the number whose square is $-1$, so it is the square root of $-1$.

$\sqrt{-1} = i$

Split off the factor of $i$, then take the real square root of $9$.

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

Build the third power from the second, using $i^2 = -1$.

$i^3 = i^2 \cdot i = (-1)\,i = -i$

Since $i^4 = 1$, one more factor of $i$ returns to the start of the cycle.

$i^5 = i^4 \cdot i = 1 \cdot i = i$

The powers repeat every four, so $i^5$ matches $i^1$.

Take the square root of both sides, keeping both signs. Both $i$ and $-i$ square to $-1$.

$x = \pm\sqrt{-1} = \pm i$

Writing only $x = i$ misses the second solution.

Replace $i^2$ with its value $-1$, then add.

$i^2 + 1 = -1 + 1 = 0$

Factor out the $i$, then take the real square root of $100$.

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

Square the $3$ and the $i$ separately, then use $i^2 = -1$.

$(3i)^2 = 9 i^2 = 9(-1) = -9$

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

$x = \pm\sqrt{-16} = \pm 4i$

Three factors of $i$ make $i^3$, which is built from $i^2 = -1$.

$i \cdot i \cdot i = i^3 = i^2 \cdot i = -i$