If the complex number a is such that |a|=1 an | vedclass

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(D) Given $|a|=1$ and $\arg (a)=	heta$,we have $a = \cos 	heta + i \sin 	heta = e^{i	heta}$.The equation is $\left(\frac{1+i z}{1-i z}ight)^4 =

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$	an \left(\frac{2 k \pi+	heta}{8}ight), k=0,1,2,3$

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(D) Given $|a|=1$ and $\arg (a)=	heta$,we have $a = \cos 	heta + i \sin 	heta = e^{i	heta}$.The equation is $\left(\frac{1+i z}{1-i z}ight)^4 = e^{i	heta}$.Taking the fourth root,$\frac{1+i z}{1-i z} = e^{i\left(\frac{2k\pi+	heta}{4}ight)} = \cos \phi + i \sin \phi$,where $\phi = \frac{2k\pi+	heta}{4}$ and $k=0,1,2,3$.Let $\omega = \frac{2k\pi+	heta}{8}$. Then $\phi = 2\omega$.Using componendo and dividendo on $\frac{1+iz}{1-iz} = \cos 2\omega + i \sin 2\omega$:$\frac{1}{iz} = \frac{\cos 2\omega + i \sin 2\omega + 1}{\cos 2\omega + i \sin 2\omega - 1} = \frac{2\cos^2 \omega + 2i \sin \omega \cos \omega}{-2\sin^2 \omega + 2i \sin \omega \cos \omega} = \frac{2\cos \omega (\cos \omega + i \sin \omega)}{2i \sin \omega (\cos \omega + i \sin \omega)} = \frac{\cos \omega}{i \sin \omega} = \frac{1}{i 	an \omega}$.Thus,$z = 	an \omega = 	an \left(\frac{2k\pi+	heta}{8}ight)$ for $k=0,1,2,3$.

If the complex number $a$ is such that $|a|=1$ and $\arg (a)=\theta$,then the roots of the equation $\left(\frac{1+i z}{1-i z}\right)^4=a$ are $z=$

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