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Sava's question via email about solving complex number equations

Prove It

Well-known member
MHB Math Helper
Jan 26, 2012
1,403
$\displaystyle \begin{align*} z^3 + 1 = 0 \end{align*}$
$\displaystyle \begin{align*} z^3 + 1 &= 0 \\ z^3 &= -1 \\ z^3 &= \mathrm{e}^{ \left( 2\,n + 1 \right) \,\pi\,\mathrm{i} } \textrm{ where } n \in \mathbf{Z} \\ z &= \left[ \mathrm{e}^{\left( 2\,n + 1 \right) \, \pi \,\mathrm{i}} \right] ^{\frac{1}{3}} \\ &= \mathrm{e}^{ \frac{\left( 2\,n + 1 \right) \,\pi}{3} \,\mathrm{i} } \end{align*}$

For the three solutions with $\displaystyle \begin{align*} \textrm{arg}\,\left( z \right) \in \left( -\pi , \pi \right] \end{align*}$, let $\displaystyle \begin{align*} n = 0 \end{align*}$ to find

$\displaystyle \begin{align*} z_1 &= \mathrm{e}^{ \frac{\pi}{3}\,\mathrm{i} } \\ &= \cos{ \left( \frac{\pi}{3} \right) } + \mathrm{i}\sin{ \left( \frac{\pi}{3} \right) } \\ &= \frac{1}{2} + \frac{\sqrt{3}}{2}\,\mathrm{i} \end{align*}$

Let $\displaystyle \begin{align*} n = 1 \end{align*}$ to find

$\displaystyle \begin{align*} z_2 &= \mathrm{e}^{ \frac{3\,\pi}{3}\,\mathrm{i} } \\ &= \mathrm{e}^{ \mathrm{\pi}\,\mathrm{i} } \\ &= \cos{ \left( \pi \right) } + \mathrm{i}\sin{ \left( \pi \right) } \\ &= -1 + 0\,\mathrm{i} \end{align*}$

If we let n be any larger, we would end up with an angle outside the acceptable range for the argument, so instead we will go the other way and let $\displaystyle \begin{align*} n = -1 \end{align*}$ to find

$\displaystyle \begin{align*} z_3 &= \mathrm{e}^{ -\frac{\pi}{3}\,\mathrm{i} } \\ &= \cos{ \left( -\frac{\pi}{3} \right) } + \mathrm{i}\sin{ \left( -\frac{\pi}{3} \right) } \\ &= \cos{ \left( \frac{\pi}{3} \right) } - \mathrm{i}\sin{\left( \frac{\pi}{3} \right) } \\ &= \frac{1}{2} - \frac{\sqrt{3}}{2}\,\mathrm{i} \end{align*}$


z^4 = 1 + \mathrm{i}
We should write this in polar form.

$\displaystyle \begin{align*} \left| z^4 \right| &= \sqrt{1^2 + 1^2} \\ &= \sqrt{1 + 1} \\ &= \sqrt{2} \end{align*}$

and as the number is in the first quadrant, that means

$\displaystyle \begin{align*} \textrm{arg}\,\left( z \right) &= \arctan{ \left( \frac{1}{1} \right) } \\ &= \arctan{ \left( 1 \right) } \\ &= \frac{\pi}{4} \end{align*}$

so we have

$\displaystyle \begin{align*} z^4 &= \sqrt{2} \,\mathrm{e}^{ \left( \frac{\pi}{4} + 2\,\pi\,n \right) \,\mathrm{i}} \textrm{ where } n \in \mathbf{Z} \\ z &= \left\{ \sqrt{2}\,\mathrm{e}^{ \left[ \frac{\left( 1 + 8\,n \right) \, \pi}{4} \right] \,\mathrm{i} } \right\} ^{ \frac{1}{4} } \\ &= \sqrt[8]{2}\,\mathrm{e}^{ \left[ \frac{\left( 1 + 8\,n \right) \,\pi }{16} \right] \,\mathrm{i} } \end{align*}$

Now to get the four roots with $\displaystyle \begin{align*} \textrm{arg}\,\left( z \right) \in \left( -\pi , \pi \right] \end{align*}$ we start by letting $\displaystyle \begin{align*} n = 0 \end{align*}$ to find

$\displaystyle \begin{align*} z_1 &= \sqrt[8]{2}\,\mathrm{e}^{ \frac{\pi}{16}\,\mathrm{i} } \\ &= \sqrt[8]{2}\,\left[ \cos{ \left( \frac{\pi}{16} \right) } + \mathrm{i}\sin{ \left( \frac{\pi}{16} \right) } \right] \\ &= \sqrt[8]{2}\cos{ \left( \frac{\pi}{16} \right) } + \sqrt[8]{2}\sin{ \left( \frac{ \pi}{16} \right) } \,\mathrm{i} \end{align*}$

Let $\displaystyle \begin{align*} n = 1 \end{align*}$ to find

$\displaystyle \begin{align*} z_2 &= \sqrt[8]{2}\,\mathrm{e}^{ \frac{9\,\pi}{16}\,\mathrm{i} } \\ &= \sqrt[8]{2}\,\left[ \cos{ \left( \frac{9\,\pi}{16} \right) } + \mathrm{i}\sin{ \left( \frac{9\,\pi}{16} \right) } \right] \\ &= \sqrt[8]{2}\cos{ \left( \frac{9\,\pi}{16} \right) } + \sqrt[8]{2}\sin{ \left( \frac{9\,\pi}{16} \right) } \,\mathrm{i} \end{align*}$

If we let n be anything greater, then we will end up with an argument outside our acceptable range, so instead let $\displaystyle \begin{align*} n = -1 \end{align*}$ to find

$\displaystyle \begin{align*} z_3 &= \sqrt[8]{2}\,\mathrm{e}^{ -\frac{7\,\pi}{16}\,\mathrm{i} } \\ &= \sqrt[8]{2}\,\left[ \cos{ \left( -\frac{7\,\pi}{16} \right) } + \mathrm{i}\sin{ \left( -\frac{7\,\pi}{16} \right) } \right] \\ &= \sqrt[8]{2}\cos{ \left( -\frac{7\,\pi}{16} \right) } + \sqrt[2]{8}\sin{ \left( -\frac{7\,\pi}{16} \right) } \, \mathrm{i} \end{align*}$

and let $\displaystyle \begin{align*} n = -2 \end{align*}$ to find

$\displaystyle \begin{align*} z_4 &= \sqrt[8]{2}\,\mathrm{e}^{ -\frac{15\,\pi}{16}\,\mathrm{i} } \\ &= \sqrt[8]{2}\,\left[ \cos{ \left( -\frac{15\,\pi}{16} \right) } + \mathrm{i}\sin{ \left( -\frac{15\,\pi}{16} \right) } \right] \\ &= \sqrt[8]{2}\cos{ \left( -\frac{15\,\pi}{16} \right) } + \sqrt[8]{2}\sin{ \left( -\frac{15\,\pi}{16} \right) } \,\mathrm{i} \end{align*}$