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

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*6(\cos(40^\circ+145^\circ)+i\sin(40^\circ+145^\circ)=42(\cos185^\circ+i\sin185^\circ)=42e^(i185^\circ)`

Answer 2

ANSWER


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

Step-by-step 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 * 6( \cos( 40 \degree +145 \degree) + i \sin(40 \degree +145 \degree ))`




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