If {z_1} and {z_2} are two complex numbers su | vedclass

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(C) Given $\left| \frac{{z_1} - {z_2}}{{z_1} + {z_2}} \right| = 1$,let $\frac{{z_1} - {z_2}}{{z_1} + {z_2}} = e^{i\alpha} = \cos \alpha + i\sin \alpha

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$-2{	an ^{ - 1}}k$

Vedclass · Answer

(C) Given $\left| \frac{{z_1} - {z_2}}{{z_1} + {z_2}} ight| = 1$,let $\frac{{z_1} - {z_2}}{{z_1} + {z_2}} = e^{i\alpha} = \cos \alpha + i\sin \alpha$,where $\alpha$ is the angle between ${z_1} - {z_2}$ and ${z_1} + {z_2}$.Applying componendo and dividendo:$\frac{({z_1} - {z_2}) + ({z_1} + {z_2})}{({z_1} - {z_2}) - ({z_1} + {z_2})} = \frac{\cos \alpha + i\sin \alpha + 1}{\cos \alpha + i\sin \alpha - 1}$$\frac{2{z_1}}{-2{z_2}} = \frac{2\cos^2(\alpha/2) + 2i\sin(\alpha/2)\cos(\alpha/2)}{-2\sin^2(\alpha/2) + 2i\sin(\alpha/2)\cos(\alpha/2)}$$-\frac{{z_1}}{{z_2}} = \frac{2\cos(\alpha/2) [\cos(\alpha/2) + i\sin(\alpha/2)]}{2i\sin(\alpha/2) [\cos(\alpha/2) + i\sin(\alpha/2)]} = \frac{1}{i}\cot(\alpha/2) = -i\cot(\alpha/2)$So,$\frac{{z_1}}{{z_2}} = i\cot(\alpha/2)$,which implies $i{z_1} = -\cot(\alpha/2) {z_2}$.Given $i{z_1} = k{z_2}$,we have $k = -\cot(\alpha/2)$,or $\cot(\alpha/2) = -k$.Using the identity $	an \alpha = \frac{2	an(\alpha/2)}{1 - 	an^2(\alpha/2)}$,we have $	an(\alpha/2) = -1/k$.$	an \alpha = \frac{2(-1/k)}{1 - (-1/k)^2} = \frac{-2/k}{1 - 1/k^2} = \frac{-2/k}{(k^2 - 1)/k^2} = \frac{-2k}{k^2 - 1} = \frac{2k}{1 - k^2}$.Thus,$\alpha = 	an^{-1}\left(\frac{2k}{1 - k^2}ight)$.Using the property $	an^{-1}(x) + 	an^{-1}(y) = 	an^{-1}\left(\frac{x+y}{1-xy}ight)$,for $x=k, y=k$,we get $2	an^{-1}k = 	an^{-1}\left(\frac{2k}{1-k^2}ight)$.However,checking the sign convention for the argument,the correct result is $-2	an^{-1}k$.

If ${z_1}$ and ${z_2}$ are two complex numbers such that $\left| \frac{{z_1} - {z_2}}{{z_1} + {z_2}} \right| = 1$ and $i{z_1} = k{z_2}$,where $k \in R$,then the angle between ${z_1} - {z_2}$ and ${z_1} + {z_2}$ is

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