The Imaginary Unit and Complex Numbers: Practice

This is the defining property of the imaginary unit, the single equation the whole chapter is built on.

$i^2 = -1$

Every other fact about complex numbers is deduced from this one equation together with the ordinary rules of arithmetic.

The imaginary unit is a new number introduced so that the previously unsolvable equation $x^2 = -1$ has a solution.

$i^2 = -1$

Note that $i$ itself is not $-1$: the number $-1$ is real, and its square is $+1$, not $-1$.

For positive $k$ the definition reads $\sqrt{-k} = i\sqrt{k}$.

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

Check by squaring: $(4i)^2 = 16\,i^2 = -16$, as required.

Apply the definition of the square root of a negative with $k = 1$.

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

The radical names the single number $i$. The equation $x^2 = -1$ has two solutions, $\pm i$, but the symbol $\sqrt{-1}$ names exactly one of them.

Compare with the standard form $a + bi$: here $a = 6$ and $b = 2$.

$\text{real part} = a = 6$

The real part is the term without the factor of $i$.

A pure imaginary number has the form $0 + bi$ with $b \ne 0$: its real part is zero and its imaginary part is not.

$7i = 0 + 7i$

The numbers $7$ and $-5$ are real, and $3 + 2i$ has both parts nonzero, so it is neither real nor pure imaginary.

Apply the definition $\sqrt{-k} = i\sqrt{k}$ with $k = 25$.

$\sqrt{-25} = i\sqrt{25} = 5i$

The radical names a single number. The two solutions $\pm 5i$ belong to the equation $x^2 = -25$, not to the symbol $\sqrt{-25}$.

Rewrite the subtraction as adding a negative coefficient.

$4 - 7i = 4 + (-7)i$

So $a = 4$ and $b = -7$. The sign travels with the imaginary part, and $b$ is the real number $-7$, never $-7i$.

Check the three cases for a real number $x$: a positive times a positive is positive, zero squared is zero, and a negative times a negative is positive.

$x^2 \ge 0 \ \text{ for every real } x$

So $x^2$ can never equal $-1$ while $x$ is real. The equation does have solutions, $\pm i$, but they live outside the real numbers.

Split off a full cycle of four.

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

After $i^4 = 1$ the powers restart, so the fifth power matches the first.

Two complex numbers in standard form are equal exactly when their parts match, so match the imaginary parts.

$b = 8$

The imaginary part is the real coefficient of $i$, so the answer is $8$, not $8i$.

Every real number is a complex number with imaginary part zero.

$12 = 12 + 0i$

So the imaginary part exists and equals $0$. That is exactly what makes the real numbers a slice of the complex numbers rather than something outside them.