Mathematics, 18.10.2019 19:40, Juicebox18
If z1 = 6(cos 3pi/2 + isin 3pi/2) and z2 = 2(cos 5pi/6 + isin 5pi/6), then the argument of z1/z2=
Answers: 3
![z1=6(cos\frac{3\pi }{2} +isin\frac{3\pi }{2}) \\z2=[tex]We are to find the argument of z1/z2Since both are in mod, argument form we can use Demoivre theorem for complex numbers to find the quotientQuotient = [tex]\frac{z1}{z2} =\frac{6(cos\frac{3\pi }{2} +isin\frac{3\pi }{2})}{2(cos\frac{5\pi }{6} +isin\frac{5\pi }{6})} \\=\frac{6}{2} (2(cos\frac{4\pi }{6} +isin\frac{4\pi }{6})\\=3 (cos\frac{2\pi }{3} +isin\frac{2\pi }{3})](/tpl/images/0332/1780/2b019.png)
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If z1 = 6(cos 3pi/2 + isin 3pi/2) and z2 = 2(cos 5pi/6 + isin 5pi/6), then the argument of z1/z2=...
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