If z=x+iy, x^2+y^2=1 and z_1=ze^{itheta}, the | vedclass

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(D) Given $z=x+iy$ and $x^2+y^2=1$, we can write $z=e^{i\phi}$ where $\phi = 	an^{-1}(\frac{y}{x})$.$z_1 = ze^{i	heta} = e^{i\phi}e^{i	heta} = e^{i

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$i 	an (n(	heta+	an^{-1} \frac{y}{x}))$

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(D) Given $z=x+iy$ and $x^2+y^2=1$, we can write $z=e^{i\phi}$ where $\phi = 	an^{-1}(\frac{y}{x})$.$z_1 = ze^{i	heta} = e^{i\phi}e^{i	heta} = e^{i(\phi+	heta)}$.Then $z_1^{2n} = e^{i2n(\phi+	heta)}$.Consider the expression $\frac{z_1^{2n}-1}{z_1^{2n}+1} = \frac{e^{i2n(\phi+	heta)}-1}{e^{i2n(\phi+	heta)}+1}$.Multiply numerator and denominator by $e^{-in(\phi+	heta)}$:$= \frac{e^{in(\phi+	heta)} - e^{-in(\phi+	heta)}}{e^{in(\phi+	heta)} + e^{-in(\phi+	heta)}} = \frac{2i \sin(n(\phi+	heta))}{2 \cos(n(\phi+	heta))}$.$= i 	an(n(\phi+	heta)) = i 	an(n(	heta+	an^{-1}(\frac{y}{x})))$.

If $z=x+iy$, $x^2+y^2=1$ and $z_1=ze^{i\theta}$, then $\frac{z_1^{2n}-1}{z_1^{2n}+1}=$

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