Question

Find the product of z1 and z2, where z1 = 7(cos 40° + i sin 40°) and z2 = 6(cos 145° + i sin 145°).

Answer 1

ANSWER

$\boxed {z_1 z_2 = 42( \cos(185 \degree) + i \sin(185 \degree ))}$

EXPLANATION

The given complex numbers are:

$z_1 = 7( \cos(40 \degree) + i \sin(40 \degree) )$


and

$z_2= 6( \cos(145 \degree) + i \sin(145\degree) )$


Recall that;

If

$z_1 = r_1 ( \cos( \theta_1) + i \sin(\theta_1 ))$

and

$z_2 = r_2 ( \cos(\theta_2) + i \sin(\theta_2 ))$

Then,


$z_1 z_2 = r_1 r_2 ( \cos( \theta_1 +\theta_2) + i \sin(\theta_1 + \theta_2 ))$



This implies that,



$z_1 z_2 = 7 \times 6( \cos( 40 \degree +145 \degree) + i \sin(40 \degree +145 \degree ))$



$z_1 z_2 = 42( \cos(185 \degree) + i \sin(185 \degree ))$

Answer 2

For two complex numbers $z_1=re^{i\theta}=r(\cos\theta+i\sin\theta)$ and $z_2=se^{i\varphi}=s(\cos\varphi+i\sin\varphi)$, the product is

$z_1z_2=rse^{i(\theta+\varphi)}=rs(\cos(\theta+\varphi)+i\sin(\theta+\varphi))$

That is, you multiply the moduli and add the arguments. You have $z_1=7e^{i40^\circ}$ and $z_2=6e^{i145^\circ}$, so the product is

$z_1z_2=7\times6(\cos(40^\circ+145^\circ)+i\sin(40^\circ+145^\circ)=42(\cos185^\circ+i\sin185^\circ)=42e^{i185^\circ}$