solve equation $z^4=7/4+6i$ complex number equation

Question

Please help me to solve the equation $z^4=(7/4)+6i$

numbers are pretty complicated so I dont know how to solve it without using calculator and with exact number.

Answer 1

Let $\omega = \dfrac{7}{4}+6i$. Then, $|\omega| = \sqrt{\left(\dfrac{7}{4}\right)^2+6^2} = \dfrac{25}{4}$ and $\arg \omega = \arctan \dfrac{6}{\frac{7}{4}} = \arctan \dfrac{24}{7}$.

So, if $z^4 = \omega$, then one solution (in the first quadrant) satisfies $|z| = |\omega|^{1/4}$ and $\arg z = \dfrac{1}{4}\arg \omega$.

Getting the magnitude is easy: $|z| = |\omega|^{1/4} = \left(\dfrac{25}{4}\right)^{1/4} = \sqrt{\dfrac{5}{2}}$.

To get the angle, let $\theta = \arg z$. Then $4 \theta = \arg \omega = \arctan \dfrac{24}{7}$. So, we have $\tan 4\theta = \dfrac{24}{7}$.

Use the tangent half angle formula twice to get $\tan 2\theta = \dfrac{3}{4}$ and $\tan \theta = \dfrac{1}{3}$.

So, $z = |z|(\cos \theta + i\sin \theta) = \sqrt{\dfrac{5}{2}}\left(\dfrac{3}{\sqrt{10}} + \dfrac{1}{\sqrt{10}}i\right) = \dfrac{3}{2}+\dfrac{1}{2}i$

Then, the other solutions are $iz$, $-z$ and $-iz$.

Answer 2

\begin{align} z^{4} = \frac{7 + 24 i }{4} \end{align} can be seen as \begin{align} z^{4} &= \sqrt{ \left(\frac{7}{4}\right)^{2} + (6)^{2} } \, e^{i \tan^{-1}(24/7)} \\ &= \left( \frac{5}{2} \right)^{2} \, e^{i \tan^{-1}(24/7)} \end{align} Now \begin{align} z &\in \left\{ \pm \sqrt{\frac{5}{2}} \, e^{\frac{i}{4} \, \tan^{-1}(24/7)}, \pm i \sqrt{\frac{5}{2}} \, e^{\frac{i}{4} \, \tan^{-1}(24/7)} \right\} \\ &\in \left\{ \pm \frac{3+i}{2}, \pm i \, \frac{3+i}{2} \right\} \\ &\in \left\{ \frac{3+i}{2}, - \frac{3+i}{2}, \frac{1-3i}{2}, - \frac{1+3 i}{2} \right\} \end{align}