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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 :-)

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

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – B. Mehta, Antonios-Alexandros Robotis, Xam, Namaste, JonMark Perry
If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ 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
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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 $y=-\frac\pi6+2k\pi$ ($k\in\mathbb Z$).

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  • $\begingroup$ 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
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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: $$\left\{2\ln2+\left(\frac{11\pi}{6}+2\pi k\right)i|k\in\{0,1,2\}\right\}.$$

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    $\begingroup$ 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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