I have the following: What are the solutions to:


in the circle around $z=0$ of radius $R=2$

Apllying the complex $LOG$

I derived: $$z^2 = \pi*i(1+2k)$$

What is the simplest way to continue from here? I tried to write $z=x+iy$ and then derived that $x=+-y$ and then: $2x^2=\pi*i(1+2k)$ but im getting mixed with positive and negetive values and the values of $k$ are unclear...how can I solve this one easily? Thanks

  • 3
    $\begingroup$ $e^z*e^z = e^{2z}$ and not $e^{z^2}$ $\endgroup$ – benji Feb 4 '15 at 0:08
  • 1
    $\begingroup$ Also one can respect convention by dropping the $*$ symbol to denote multiplication; we are not discussing programming languages. $\endgroup$ – P Vanchinathan Feb 4 '15 at 0:42

One way would be to put things in polar form: write $z=r e^{i \theta}$, note that $i=e^{i \pi/2}$, and figure out what r and $\theta$ are.

EDIT: Didn't actually read your whole question. This would be how to solve $z^2 = (2 k + 1) \pi i$, but it's overkill for your actual question, which is how to solve $e^{2z} = -1$.

  • $\begingroup$ If im going for polar cords, im getting: $R^{2}e^{2\theta i}=e^{\pi /2i+2\pi m}(\pi+2k\pi)$ but now im getting just one root inside the circle radius 2 for $k=0$ and $m=0$ and I know the answer should be 4 roots $\endgroup$ – user3921 Feb 4 '15 at 7:22
  • $\begingroup$ Well, as it's been pointed out above, you're not looking for solutions to $z^2 = (2k+1)\pi i$. $\endgroup$ – Daniel McLaury Feb 4 '15 at 20:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.