Let z1 = √2(cos 7pi/4 + i sin 7pi/4) and z2 = 2(cos 5pi/3 + i sin 5pi/3), how do you find z1/z2?

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Let z1 = √2(cos 7pi/4 + i sin 7pi/4) and z2 = 2(cos 5pi/3 + i sin 5pi/3), how do you find z1/z2?

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Answer 1 · 2016-04-13T07:00:44

$\frac{z_{1}}{z_{2}} = \frac{1}{\sqrt{2}} [cos (\frac{π}{12}) + i sin (\frac{π}{12})]$ $z_1/z_2=1/sqrt2[cos(pi/12)+isin(pi/12)]$

Explanation:

A complex number of form $r cos θ + i r sin θ$ $rcostheta+irsintheta$ can also be written as $r \cdot e^{i θ}$ $r*e^(itheta)$ .

Hence, $z_{1} = \sqrt{2} (cos (7 \frac{π}{4}) + i sin (7 \frac{π}{4})) = \sqrt{2} e^{i 7 \frac{π}{4}}$ $z_1=sqrt2(cos(7pi/4)+isin(7pi/4))=sqrt2e^(i7pi/4)$

and $z_{2} = 2 (cos (5 \frac{π}{3}) + i sin (5 \frac{π}{2})) = 2 e^{i 5 \frac{π}{3}}$ $z_2=2(cos(5pi/3)+isin(5pi/2))=2e^(i5pi/3)$

Hence $\frac{z_{1}}{z_{2}} = \frac{\sqrt{2}}{2} x e^{i (7 \frac{π}{4} - 5 \frac{π}{3})} = \frac{1}{\sqrt{2}} \cdot e^{i (\frac{21 π - 20 π}{12})} = \frac{1}{\sqrt{2}} \cdot e^{i \frac{π}{12}}$ $z_1/z_2=sqrt2/2xe^(i(7pi/4-5pi/3))=1/sqrt2*e^(i((21pi-20pi)/12))=1/sqrt2*e^(i(pi)/12)$

= $\frac{1}{\sqrt{2}} [cos (\frac{π}{12}) + i sin (\frac{π}{12})]$ $1/sqrt2[cos(pi/12)+isin(pi/12)]$

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Let z1 = √2(cos 7pi/4 + i sin 7pi/4) and z2 = 2(cos 5pi/3 + i sin 5pi/3), how do you find z1/z2?

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