Mathematics-Online course: Preparatory Course Mathematics-Basics-Complex Numbers-Powers of Complex Numbers

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	Mathematics-Online course: Preparatory Course Mathematics - Basics - Complex Numbers
	Powers of Complex Numbers

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For $z = r e^{\mathrm{i}\varphi}$ and a rational exponent ,

$\displaystyle z^{p/q} = r^{p/q} \exp(p(\varphi+2\pi j)\mathrm{i}/q),\quad j=0,\ldots,q-1 \,.$

The fractional power is ambiguous since $\varphi=\operatorname{arg}(z)$ is determined only up to multiples of $2\pi$ . Possible values differ by powers of the

-th roots of unity:

$\displaystyle w_q^0,w_q^p,\ldots,w_q^{(q-1)p},\quad w_q = \exp(2\pi\mathrm{i}/q) \,.$

(Authors: Höllig/Hörner/Kopf/Abele)

In order to calculate

$\displaystyle z =(-1+\mathrm{i})^{2/3}$

we transform

into polar coordinates. This yields

$\displaystyle \left(\sqrt{2} \exp\left(\frac{3 \pi \mathrm{i}}{4}\right)\right)... ...sqrt[3]{2} \exp\left(\frac{\mathrm{i}\pi}{2}\right) w_3^{2k}, \quad k=0,1,2\,,$

with $w_3 = \exp(2\pi\mathrm{i}/3)$ . Thus, possible values are

$\displaystyle z_0 =$	$\displaystyle \sqrt[3]{2}\; \mathrm{i} \,,$
$\displaystyle z_1 =$	$\displaystyle \sqrt[3]{2} \; \mathrm{i} \exp\left( \frac{4 \pi \mathrm{i}}{3} \right) = \sqrt[3]{2}\left( \frac{\sqrt{3}}{2}-\frac{\mathrm{i}}{2} \right)\,,$
$\displaystyle z_2 =$	$\displaystyle \sqrt[3]{2}\; \mathrm{i} \exp\left( \frac{8 \pi \mathrm{i}}{3} \right) = \sqrt[3]{2}\left( -\frac{\sqrt{3}}{2} -\frac{\mathrm{i}}{2} \right)\,.$

The results can be easily checked. For example, we have

$\displaystyle z_1^3 = \left(\sqrt[3]{2} \;\mathrm{i} \exp\left( \frac{4 \pi \mathrm{i}}{3} \right) \right)^3 = -2 \,\mathrm{i} \,,$

which equals $(-1+\mathrm{i})^2$ .

For irrational and purely imaginary exponents, one generally obtains infinitely many solutions, as shown in the following examples:

infinitely many solutions on the unit circle:

$\displaystyle \mathrm{i}^{\pi}$ $\displaystyle =$ $\displaystyle \exp ((\pi/2+2\pi k)\mathrm{i})^{\pi}$

$\displaystyle =$ $\displaystyle \exp \left(\mathrm{i}[\pi^2/2+2\pi^2 k]\right),\qquad k \in \mathbb{Z}$
infinitely many solutions on a half line:

$\displaystyle \pi^{\mathrm{i}}$ $\displaystyle =$ $\displaystyle \exp (\ln \pi + 2\pi k\mathrm{i})^{\mathrm{i}} = \exp (\mathrm{i} \ln \pi -2\pi k)$

$\displaystyle =$ $\displaystyle \exp(-2\pi k) \exp(\mathrm{i}\ln \pi), \qquad k \in \mathbb{Z}$
infinitely many solutions on the positive real line:

$\displaystyle \mathrm{i}^{\mathrm{i}}$ $\displaystyle =$ $\displaystyle \exp ((\pi/2+2\pi k)\mathrm{i})^{\mathrm{i}}$

$\displaystyle =$ $\displaystyle \exp (-\pi/2-2\pi k), \qquad k \in \mathbb{Z}$

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automatically generated 1/9/2017

$\displaystyle \mathrm{i}^{\pi}$	$\displaystyle =$	$\displaystyle \exp ((\pi/2+2\pi k)\mathrm{i})^{\pi}$
	$\displaystyle =$	$\displaystyle \exp \left(\mathrm{i}[\pi^2/2+2\pi^2 k]\right),\qquad k \in \mathbb{Z}$

$\displaystyle \pi^{\mathrm{i}}$	$\displaystyle =$	$\displaystyle \exp (\ln \pi + 2\pi k\mathrm{i})^{\mathrm{i}} = \exp (\mathrm{i} \ln \pi -2\pi k)$
	$\displaystyle =$	$\displaystyle \exp(-2\pi k) \exp(\mathrm{i}\ln \pi), \qquad k \in \mathbb{Z}$

$\displaystyle \mathrm{i}^{\mathrm{i}}$	$\displaystyle =$	$\displaystyle \exp ((\pi/2+2\pi k)\mathrm{i})^{\mathrm{i}}$
	$\displaystyle =$	$\displaystyle \exp (-\pi/2-2\pi k), \qquad k \in \mathbb{Z}$