Let $z_{1} = 2 [cos (\frac{π}{6}) + i sin (\frac{π}{6})]$ $z_1=2[cos(pi/6)+isin(pi/6)]$ How do you find $\frac{1}{z_{1}} ?$ $1/z_1?$

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Answer 1 · 2016-12-05T09:25:28

$\frac{1}{z_{1}} = \frac{\sqrt{3}}{4} - i \frac{1}{4}$ $1/z_1=sqrt3/4-i1/4$

$\frac{1}{z_{1}} = z_{1}^{- 1}$ $1/(z_1)=z_1^-1$

using De'Moivre's theorem

$\frac{1}{z_{1}} = z_{1}^{- 1} = {[2 (cos (\frac{π}{6}) + i sin (\frac{π}{6}))]}^{- 1}$ $1/z_1=z_1^-1=[2(cos(pi/6)+isin(pi/6))]^-1$

$\frac{1}{z_{1}} = z_{1}^{- 1} = [2^{- 1} (cos (- \frac{π}{6}) + i sin (- \frac{π}{6}))]$ $1/z_1=z_1^-1=[2^-1(cos(-pi/6)+isin(-pi/6))]$

$cos θ$ $costheta$ is an even function; $sin θ$ $""sintheta$ an odd one#

$\frac{1}{z_{1}} = z_{1}^{- 1} = [\frac{1}{2} (cos (\frac{π}{6}) - i sin (\frac{π}{6}))]$ $1/z_1=z_1^-1=[1/2(cos(pi/6)-isin(pi/6))]$

$\frac{1}{z_{1}} = z_{1}^{- 1} = [\frac{1}{2} (\frac{\sqrt{3}}{2} - i \frac{1}{2})]$ $1/z_1=z_1^-1=[1/2(sqrt3/2-i1/2)]$

$\frac{1}{z_{1}} = \frac{\sqrt{3}}{4} - i \frac{1}{4}$ $1/z_1=sqrt3/4-i1/4$

Let z1=2[cos(π6)+isin(π6)]z_1=2[cos(pi/6)+isin(pi/6)] How do you find 1z1?1/z_1?