$z_1= \cos(4\pi/3) + i\sin(4\pi/3)$
$z_2= \cos(π/3) + i\sin (π/3)$
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.
On
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:
$$z_1=-\frac12\left(1+\sqrt3\,i\right)\;,\;\;z_2=\frac12\left(1+\sqrt3\,i\right)\;\ldots$$
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.