I have been working with this equation for a long time now, and I just can't figure it out.

Solve the following equation. $$e^z=2\sqrt3 - 2i.$$

There should be three solutions that have a positive imaginary part. I hope some of you, can help me. Thank you :-)


closed as off-topic by B. Mehta, Antonios-Alexandros Robotis, Xam, Namaste, JonMark Perry Sep 24 '17 at 0:09

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  • $\begingroup$ Write the right side as $e^w$ for some $w\in \mathbb{C}$. $\endgroup$ – anomaly Sep 23 '17 at 8:40
  • $\begingroup$ Here's a MathJax tutorial :) $\endgroup$ – Shaun Sep 23 '17 at 8:42

If $z=x+yi$, then $e^z=e^x\bigl(\cos y+i\sin y\bigr)$. So$$e^z=2\sqrt3-2i=4\left(\frac{\sqrt3}2-\frac12i\right)\iff e^x=4\wedge\cos y=\frac{\sqrt3}2\wedge\sin y=-\frac12.$$So, take $x=\log4$, and $y=-\frac\pi6+2k\pi$ ($k\in\mathbb Z$).

  • $\begingroup$ My edit was for style. I thought the first line didn't need so many brackets......+1 $\endgroup$ – DanielWainfleet Sep 23 '17 at 9:03
  • $\begingroup$ So to get the three solutions with positive imaginary parts - should I then "just" put in some random constants instead of k? Example: log(4)-Pi/6 +2·7·Pi $\endgroup$ – Line Sep 23 '17 at 9:06
  • $\begingroup$ @Caroline The example that you provided is not a solution. But, yes, you should choose $k\in\mathbb Z$ such that $y>0$. $\endgroup$ – José Carlos Santos Sep 23 '17 at 9:16

Let $z=x+yi$, where $\{x,y\}\subset\mathbb R$.

Thus, $$e^x(\cos{y}+i\sin{y})=2\sqrt3-2i,$$ which gives $$e^x\cos{y}=2\sqrt3$$ and $$e^x\sin{y}=-2.$$

Thus, $$e^{2x}=(2\sqrt3)^2+(-2)^2,$$ which gives $$e^{2x}=16$$ or $$x=2\ln2.$$ Hence, $\cos{y}=\frac{\sqrt3}{2}$ and $\sin{y}=-\frac{1}{2}$, which gives $$y=\frac{11\pi}{6}+2\pi k,$$ where $k\in\mathbb Z$.

Id est, we got the answer: $$\left\{2\ln2+\left(\frac{11\pi}{6}+2\pi k\right)i|k\in\mathbb Z\right\}.$$ If you wish three solutions with positive imaginary part then we obtain for example: $$\left\{2\ln2+\left(\frac{11\pi}{6}+2\pi k\right)i|k\in\{0,1,2\}\right\}.$$

  • 1
    $\begingroup$ Use radians when computing the arguments of complex numbers. That is, $330^\circ=\frac{11\pi}6$. $\endgroup$ – robjohn Sep 23 '17 at 9:12

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