# Calculators and complex numbers (Part 18)

In this series of posts, I explore properties of complex numbers that explain some surprising answers to exponential and logarithmic problems using a calculator (see video at the bottom of this post). These posts form the basis for a sequence of lectures given to my future secondary teachers.

To begin, we recall that the trigonometric form of a complex number $z = a+bi$ is $z = r(\cos \theta + i \sin \theta) = r e^{i \theta}$

where $r = |z| = \sqrt{a^2 + b^2}$ and $\tan \theta = b/a$, with $\theta$ in the appropriate quadrant. As noted before, this is analogous to converting from rectangular coordinates to polar coordinates.

Definition. If $z$ is a complex number, then we define $e^z = \displaystyle \sum_{n=0}^{\infty} \frac{z^n}{n!}$

This of course matches the Taylor expansion of $e^x$ for real numbers $x$.

In the last few posts, we proved the following theorem.

Theorem. If $z$ and $w$ are complex numbers, then $e^z e^w = e^{z+w}$.

This theorem allows us to compute $e^z$ without directly plugging into the above infinite series.

Theorem. If $z = x + i y$, where $x$ and $y$ are real numbers, then $e^z = e^x (\cos y + i \sin y)$

Proof. With the machinery that’s been developed over the past few posts, this one is actually a one-liner: $e^z = e^{x+iy} = e^x e^{iy} = e^x (\cos y + i \sin y)$.

For example, $e^{4+\pi i} = e^4 (\cos \pi + i \sin \pi) = -e^4$

Notice that, with complex numbers, it’s perfectly possible to take $e$ to a power and get a negative number. Obviously, this is impossible when using only real numbers.

Another example: $e^{-2+3i} = e^{-2} (\cos 3 +i \sin 3)$

In this answer, we have to remember that the angle is 3 radians and not 3 degrees. For completeness, here’s the movie that I use to engage my students when I begin this sequence of lectures.

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