Complex number multiplication

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

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.

• Do you know about the polar form of complex numbers? – 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.

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

Hints:

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

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

$$z_1=\cos\frac{4\pi}3+i\sin\frac{4\pi}3=e^{i\frac{4\pi}3}\;,\;\;z_2=e^{i\frac\pi3}\implies$$

$$z_1z_2=e^{\frac{5\pi}3i}$$

Or even in rectangular form:

$$z_1=-\frac12\left(1+\sqrt3\,i\right)\;,\;\;z_2=\frac12\left(1+\sqrt3\,i\right)\;\ldots$$

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