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Answered

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20 POINTS
Using mathematically precise language, explain in detail how you would multiply the complex number
z1=r1 (cos theta1+isin theta 1)
with the complex number
z2=r2 (cos theta 2+isin theta 2)

r-radius/radii
m-modulus/moduli
theta-argument represent angle theta

Sagot :

The product  complex number is  z1z2 = r1r2[cos (θ1+θ2) + isin (θ1+θ2)]

[tex]Z_1\times Z_2= r_1(cos \theta_1 + isin \theta_1) \times r_2(cos \theta_2 + isin \theta_2)[/tex]

r-radius/radii

m-modulus/moduli

theta-argument represent angle theta

[tex]= r_1r_2(cos \theta_1 + i sin \theta_1) (cos \theta_2 + isin \theta_2)[/tex]

[tex]= r_1r_2(cos \theta_1 cos \theta_2 + i cos \theta_1 sin \theta_2 + i sin \theta_1 cos \theta_2 + i^2sin\theta_1sin\theta_2[/tex]

What is the value of i square in complex number?

The value of i square is

[tex]i^2=-1[/tex]

Therefore,we get,

[tex]= r_1r_2(cos \theta_1 cos \theta_2 + i cos \theta_1 sin \theta_2 + i sin \theta_1 cos \theta_2 + (-1)sin \theta_1 sin \theta_2)[/tex]

[tex]= r_1r_2(cos \theta_1 cos \theta_2 + i cos \theta_1 sin \theta_2 + i sin \theta_1 cos \theta_2 - sin \theta_1 sin \theta_2)[/tex]

Since we know that,

[tex]cos(\theta_1+\theta_2)=cos\theta_1cos\theta_2-sin\theta_1sin\theta_2[/tex]

[tex]sin(\theta_1+\theta_2)=sin\theta_1cos\theta_2+cos\theta_1sin\theta_2[/tex]

So using above value we get,

[tex]= r_1r_2(cos \theta_1 cos \theta_2 - sin \theta_1 sin \theta_2) + i (cos \theta_1 sin \theta_2 + sin \theta_1 cos \theta_2)[/tex]

[tex]=r_1r_2cos(\theta_1+\theta_2)+isin(\theta_1+\theta_2)[/tex]

Therfore the product of the complex number is z1z2 = r1r2[cos (θ1+θ2) + isin (θ1+θ2)].

To learn more about the product of complex number visit:

https://brainly.com/question/1462345

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