Complex numbers find wide application in mathematics, physics and engineering. We hope the exercises and solutions included here help you solidify your understanding of complex arithmetic.

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Problems

Solutions

Online resources

Problems

#### Complex roots:

- What are the fourth roots of 1?
- Calculate the cube roots of \(i\).
- What is the square root of \(1+2i~\)?
- How many distinct nth roots of unity are there? (n is some integer).

Solutions

- For any integer k, \(~1 ~=~ e^{2\pi i ~k}~\). Thus, \(~e^{\frac{2\pi}{4}i\times 0}~,~ e^{\frac{2\pi}{4}i\times 1 }~,~ e^{\frac{2\pi}{4}i\times 2}~,~ e^{\frac{2\pi}{4}i \times 3}~\) are all fourth roots of 1. Note that \(~e^{\frac{2\pi}{4}i\times 4} ~=~ e^{\frac{2\pi}{4}i\times 0}\) , \(e^{\frac{2\pi}{4}i\times 5} ~=~ e^{\frac{2\pi}{4}i\times 1}\) and so on; thus we don’t get any new roots from other values of \(k\).
- For any integer k, \(~i ~=~ e^{\frac{\pi}{2}i ~+~ 2\pi i~k}~\). Thus, \(~e^{\frac{\pi}{6}i ~+~\frac{2\pi}{3}i\times 0}~,~ e^{\frac{\pi}{6}i ~+~\frac{2\pi}{3}i\times 1 }~,~ e^{\frac{\pi}{6}i ~+~\frac{2\pi}{3}i\times 2}~\) are all cube roots of \(i\). Similarly to the previous problem, \(e^{\frac{2\pi}{3}i\times 3} ~=~ e^{\frac{2\pi}{3}i\times 0}\) and so on; thus we don’t get any new roots from other values of \(k\).
- Let’s write \(1 + 2i\) in polar form. Note that, \(|1+2i| = \sqrt{1^2 + 2^2} ~=~ \sqrt{5}\) and \(\textrm{arg}(1+2i) ~ \equiv~ \theta ~=~ \textrm{tan}^{-1}(2) ~\sim~ 1.1\) rad. Thus, for any integer \(k\), \(1+2i ~=~ \sqrt[4]{5}~e^{i\theta ~+~ 2\pi i k}\) and so \(\sqrt[4]{5}~e^{i\frac{\theta}{2} ~+~ \frac{2\pi}{2} i \times 0}\) , \(\sqrt[4]{5}~e^{i\frac{\theta}{2} ~+~ \frac{2\pi}{2} i \times 1}\) are both square roots of \(1+2i\). Similarly to the previous solutions, all other values of \(k\) yield one of the same two roots.
- There are \(n\) nth roots of 1. For any integer k, \(1 ~=~ e^{2\pi i ~k}\). The distinct roots will be of the form, \(e^{\frac{2\pi}{n}i\times k}\) for all \(k\) from \(0\) to \(n-1\). For any other k, \(\frac{2\pi}{n} k = 2\pi*q ~+~ \frac{2\pi}{n}r\), for some integers \(q\) and \(r\) where \(0 ~\leq~ r ~\leq~ n-1\) and thus \(e^{\frac{2\pi}{n}i\times k}\) will not yield new roots.

Online resources

This short video by Prof Sam Gralla has a straightforward discussion of calculating roots of complex numbers.

I have taught physics at levels ranging from introductory classical mechanics to advanced graduate quantum mechanics, along with calculus and linear algebra.