$z_1= \cos(4\pi/3) + i\sin(4\pi/3)$
$z_2= \cos(π/3) + i\sin (π/3)$

I want to find out $z_1z_2$.

I know that $(x +iy)(u + iv$) = $(xu - yv) + i(xu + yv)$

So I want to simplify $\cos(4π/3)\cos(π/3) - \sin(4π/3)\sin(π/3)] + i[\cos(4π/3)\cos(π/3) + \sin(4π/3)\sin(π/3)]$

First of, is my formula correct and second, how would I multiply sin or cos functions?

Any assistance would be greatly appreciated.

  • $\begingroup$ Do you know about the polar form of complex numbers? $\endgroup$ – DonAntonio Mar 15 '14 at 11:57

Well you can simplify using the trigonometric identities $\cos(A \pm B) = \cos A\cos B \mp \sin A \sin B \;\;$ and $ \;\; \sin A \cos B \pm \cos A \sin B = \sin (A \pm B) $. For more trigonometric identities check this out.

So the expression reduces to $\cos \left({\frac {5\pi}{3}}\right) + i\sin \left( {\frac{ 5\pi}{3}}\right)$.

But these relationships have been generalised to apply to all multiplications of complex numbers represented in this form. Check this out.

  • $\begingroup$ Your answer is wrong. OP made a multiplication error. You didn't check and continued on that error. $\endgroup$ – Guy Mar 15 '14 at 12:05
  • 1
    $\begingroup$ Correct answer is $\cos (\frac {5\pi}{3}) + i\sin (\frac{5\pi}{3})$ $\endgroup$ – Guy Mar 15 '14 at 12:05
  • $\begingroup$ @Sabyasachi: Totally my bad. Won't happen again. Edited. $\endgroup$ – Ishfaaq Mar 15 '14 at 15:12
  • $\begingroup$ And my downvote becomes an upvote :) $\endgroup$ – Guy Mar 15 '14 at 15:15


$$\cos(a\pm b)=\cos a\cos b\mp\sin a\sin b$$

But if you know the polar form, then it is way easier:



Or even in rectangular form:


  • 1
    $\begingroup$ You should probably point out he made a sign error in the multiplication. $\endgroup$ – Guy Mar 15 '14 at 12:02
  • 1
    $\begingroup$ I don't usually read the development of questions when not written with LaTeX...and this wasn't when I first read it. $\endgroup$ – DonAntonio Mar 15 '14 at 12:03
  • 1
    $\begingroup$ Yeah I know, I edited it then. $\endgroup$ – Guy Mar 15 '14 at 12:04

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.