Now that we have seen two forms for specifying complex numbers, let us summarize the ideas for converting from one form to the other.
Let \(z = a+bi\) be a complex number. We want to write \(z\) as \(r\operatorname{cis} \theta\) with \(r = |z|= \sqrt{a^2+b^2}\) and \( 0 \leq \theta \lt 2\pi\).
We take care of a few special cases first. If \(z = 0\), then \(r = 0\). Hence, \(\theta\) can be any value. However, we will adopt the convention that \(\theta = 0\) in this case unless otherwise stated.
Suppose that \(a = 0\). In this case, the modulus is just \(r = |b|\) and \(\theta =\left\{\begin{array}{ll} \frac{\pi}{2} & \text{if } b \gt 0 \\ \frac{3\pi}{2} & \text{otherwise.} \end{array}\right.\)
Suppose that \(a \neq 0\). Let \(\alpha = \displaystyle\tan^{-1} \frac{|b|}{|a|}\). The following table shows the value of the argument \(\theta\) depending on the signs of \(a\) and \(b\):
Case | \(\theta\) |
---|---|
\(a \gt 0, b \geq 0 \) | \(\alpha\) |
\(a \lt 0, b \geq 0 \) | \(\pi - \alpha\) |
\(a \lt 0, b \lt 0 \) | \(\pi + \alpha\) |
\(a \gt 0, b \lt 0 \) | \(2\pi -\alpha\) |
To convert \(13 + 4i\) to polar form, note that both its real part and imaginary part are positive. Hence, its polar form is \(r \operatorname{cis} \theta\) where \(r = \sqrt{ 13^2 + 4^2} = \sqrt{ 285 }\) and \(\theta = \tan^{-1} \frac{4}{13}\).
To convert \(-1 - 17i\) to polar form, note that both its real part and its imaginary part are negative. Hence, its polar form is \(r \operatorname{cis} \theta\) where \(r = \sqrt{ (-1)^2 + (-17)^2} = \sqrt{ 290 }\) and \(\theta = \pi + \tan^{-1} \frac{|-17|}{|-1|} = \pi + \tan^{-1} 17\).
Let \(z = r \operatorname{cis} \theta\) be a complex number with \(r \geq 0\) and \(0 \leq \theta \lt 2\pi\).
To convert \(z\) to rectangular form, recall that \(\operatorname{cis} \theta\) is an abbreviation for \(\cos \theta + i \sin \theta\). Thus, \[z = r (\cos \theta + i\sin \theta) = (r \cos \theta) + (r \sin \theta) i.\]
In short, \(\mathbf{Re} (z) = r \cos \theta \) and \(\mathbf{Im} (z) = r \sin \theta.\)
The rectangular form of \(3 \operatorname{cis} \frac{3\pi}{2}\) is \(3 \cos \frac{3\pi}{2} + 3\sin \frac{3\pi}{2} i = 0 + (-3)i = -3i\).
The rectangular form of \(\sqrt{2} \operatorname{cis} \frac{3}{7}\) is \(\sqrt{2} \cos \frac{3}{7} + \sqrt{2}\sin \frac{3}{7} i \approx 1.2863 + 0.5877 i\) with real and imaginary parts rounded to 4 decimal places.
Convert each of the following to rectangular form with real and imaginary parts rounded to 4 decimal places.
\(\operatorname{cis} \frac{3\pi}{2}\)
\(\sqrt{3} \operatorname{cis} \frac{5\pi}{6}\)
\(2.5 \operatorname{cis} \frac{13}{5}\)
Convert each of the following to polar form.
\( 12 + 7i \)
\( -7i \)
\(7 - 12i\)
\(13 - i\)
\(-23 - 4i\)