I'm having a little trouble with this problem.

It's asking to find all $z\in\mathbb C$ that satisfy $z^3 = -2(1+i\sqrt{3})\overline z$, and to keep the answers in standard form.

I tried expanding and simplifying both sides by using $z=x+yi$ where $x,y\in\mathbb R$, then equating the imaginary and real parts separately to get:

real parts: $x^3-3xy^2 = -2x + 2\sqrt{3}y$.

imaginary parts: $4x^2y-y^3=-2\sqrt{3}x-2y$

But I don't really see how I can get very far with these two expressions because of the degree of the variables.

Am I using the wrong technique?


Try converting to polar, i.e. write $z=re^{\theta}$ and $-2(1+i\sqrt{3})=-4e^{i\pi/3}$. Then solve for $r$ and $\theta$. And then convert back to standard form.

  • $\begingroup$ But how does that account for the $\overline z$? $\endgroup$ – E 4 6 Nov 21 '14 at 0:30
  • $\begingroup$ Take the absolute value of both sides. Then you get $r^3= 4r$. $\endgroup$ – Avi Steiner Nov 21 '14 at 13:08

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.