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

2 Answers

6 votes
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)
User Baby
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2 votes
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 ))


User MarbleMunkey
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