Complex functions and polynomials: Complex functions
De Moivre's theorem
One of the most important properties of complex numbers stems from the so-called de Moivre’s theorem (or de Moivre’s formula). As we will see, it greatly simplifies the computation of integer powers of complex numbers.
Let #\green{x}# be a real number and #\orange{n}# an integer. Then, de Moivre's theorem states that \[\left(\cos(\green{x}) + \sin(\green{x})\cdot\complexi\right)^{\orange{n}}=\cos(\orange{n}\cdot \green{x})+\sin(\orange{n}\cdot \green{x})\cdot \complexi\]
With de Moivre's theorem, computing integer powers of complex numbers can be very simple when they are written in the correct form. We first look at how this looks for the polar and modulus-argument forms.
Powers of complex numbers (polar and modulus-argument form)
Let #\blue{z}# be a complex number with polar coordinates #\left[\purple{r},\green{\theta}\right]#. Then, according to de Moivre's theorem, #\blue{z}^{\orange{n}}# is given by
\[\blue{z}^{\orange{n}}
=\left(\purple{r}\cdot\left(\cos\left(\green{\theta}\right)+\sin\left(\green{\theta}\right)\cdot\ii\right)\right)^{\orange{n}}
=\purple{r}^{\orange{n}}\cdot\left(\cos\left(\orange{n}\cdot\green{\theta}\right)+\sin\left(\orange{n}\cdot\green{\theta}\right)\cdot\ii\right)\]
Let #\blue{z}# be a complex number with modulus #\purple{r}# and principal value #\green{\varphi}#. Then, according to de Moivre's theorem, #\blue{z}^{\orange{n}}# can be obtained by
\[\blue{z}^{\orange{n}}
=\left(\purple{r}\cdot\left(\cos\left(\green{\varphi}\right)+\sin\left(\green{\varphi}\right)\cdot\ii\right)\right)^{\orange{n}}
=\purple{r}^{\orange{n}}\cdot\left(\cos\left(\orange{n}\cdot\green{\varphi}\right)+\sin\left(\orange{n}\cdot\green{\varphi}\right)\cdot\ii\right)\]
Note that it is necessary to check if #-\pi<\orange{n}\cdot\green{\varphi}\leq\pi# after the computation. If this is not the case, we need to add or subtract multiples of #2\cdot \pi# until the principal value of #\blue{z}^{\orange{n}}# is in the allowed range.
We can obtain the same computational rules for the polar-exponential form in an even more compact way.
Powers of complex numbers (polar-exponential form)
Let #\blue{z}# be a complex number whose polar-exponential form is #\blue{z}=\purple{r}\cdot\e^{\green{\theta}\cdot \ii}#. According to de Moivre's theorem, #\blue{z}^{\orange{n}}# is given by
\[\blue{z}^{\orange{n}}=\purple{r}^{\orange{n}}\cdot\e^{\orange{n}\cdot\green{\theta}\cdot \ii}\]
If we use the principal value instead, we obtain
\[\blue{z}^{\orange{n}}=\purple{r}^{\orange{n}}\cdot\e^{\orange{n}\cdot\green{\varphi}\cdot \ii}\]where we once again need to ensure that #-\pi<\orange{n}\cdot\green{\varphi}\leq\pi# and add or subtract the necessary multiples of #2\cdot\pi# if that is not the case.
Examples
\[\begin{array}{rcl}
\left( \purple{\frac{1}{2}}\cdot\e^{\green{\frac{\pi}{3}}\cdot\ii} \right)^{\orange{5}}&=&\left(\purple{\frac{1}{2}}\right)^{\orange{5}}\cdot\e^{\orange{5}\cdot \green{\frac{\pi}{3}}\cdot\ii}\\
&=&\frac{1}{32}\cdot \e^{\frac{5\cdot\pi}{3}\cdot \ii}
\end{array}\]
Using principal value
\[\begin{array}{rcl}
\left(\purple{\frac{1}{2}}\cdot\e^{\green{\frac{\pi}{3}}\cdot\ii} \right)^{\orange{4}}&=&\purple{3}^{\orange{4}}\cdot\e^{\orange{4}\cdot\green{\frac{\pi}{4}}\cdot\ii}\\
&=&81\cdot\e^{\pi\cdot\ii}
\end{array}\]
Note that #\orange{n}\cdot\green{\varphi}=\orange{4}\cdot\green{\frac{\pi}{4}}=\pi#, which is within the range of the principal value.
We use de Moivres theorem: given #z=r \cdot \e^{\theta \cdot \ii}#, then #z^n=r^n \cdot \e^{n \cdot \theta \cdot \ii}#
This implies
\[ \begin{array}{rcl} \left(5 \cdot \e^{{{-\pi}\over{4}} \cdot \ii}\right)^{5}&=&5^{5}
\cdot \e^{5 \cdot {{-\pi}\over{4}} \cdot \ii} \\
&&\qquad \blue{\text{used de Moivre's theorem with }r=5 \text{, }\theta={{-\pi}\over{4}} \text{, and }n=5} \\
&=& 3125 \cdot \e^{{{-5\cdot \pi}\over{4}} \cdot \ii}\\
&&\qquad \blue{\text{computed and simplified}}
\end{array}\]
Or visit omptest.org if jou are taking an OMPT exam.
