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De Moivre's Formula

Let \(z\) be a complex number given in polar form: \(r \operatorname{cis} \theta\). Let \(n\) be an integer. De Moivre's Formula states that \[z^n = r^n \operatorname{cis} (n\theta).\] This formula simplifies computing powers of a complex number provided one has its polar form.


Let \(z = 1+i\). What is \(z^{10}\)?

There is nothing stopping us from computing \(z^{10}\) directly. However, we will use De Moivre's formula. First, we need to convert \(z\) to polar form. Note that \(r = |~z~| = \sqrt{2}\) and \(\arg (z) = \arctan(1) = \frac{\pi}{4}\). Hence, \(z = \sqrt{2} \operatorname{cis} \frac{\pi}{4}\), giving \[\begin{eqnarray} z^{10} & = & \sqrt{2}^{10}\operatorname{cis} \left(10 \cdot \frac{\pi}{4}\right) \\ & = & 32 \operatorname{cis} \frac{5\pi}{2} \\ & = & 32 \operatorname{cis} \left(2\pi + \frac{\pi}{2} \right) \\ & = & 32 \operatorname{cis} \frac{\pi}{2} \\ & = & 32 i \end{eqnarray} \]

In the above, we used the fact that \(\cos (2k\pi + \theta) = \cos \theta\) and \(\sin (2k\pi + \theta) = \sin \theta\) for all integers \(k\).

In general, if the argument is outside the interval \([0,2\pi)\), it is customary to reduce it to a value within this interval as we have done above.

Quick Quiz


  1. Compute \((\sqrt{3}-i)^{12}\).  

  2. Compute \((-1+i)^{8}\).