left(frac{cos theta+i sin theta}{sin theta+i | vedclass

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Question

(B) Let $z_1 = \frac{\cos 	heta+i \sin 	heta}{\sin 	heta+i \cos 	heta} = \frac{\cos 	heta+i \sin 	heta}{i(\cos 	heta-i \sin 	heta)} = \frac{1}

Vedclass · Accepted Answer

$2 \cos 16 	heta$

Vedclass · Answer

(B) Let $z_1 = \frac{\cos 	heta+i \sin 	heta}{\sin 	heta+i \cos 	heta} = \frac{\cos 	heta+i \sin 	heta}{i(\cos 	heta-i \sin 	heta)} = \frac{1}{i} \cdot \frac{e^{i 	heta}}{e^{-i 	heta}} = -i e^{i 2 	heta}$.Then $z_1^8 = (-i)^8 (e^{i 2 	heta})^8 = 1 \cdot e^{i 16 	heta} = \cos 16 	heta + i \sin 16 	heta$.Let $z_2 = \frac{1+\cos 	heta-i \sin 	heta}{1+\cos 	heta+i \sin 	heta} = \frac{2 \cos^2 \frac{	heta}{2} - i 2 \sin \frac{	heta}{2} \cos \frac{	heta}{2}}{2 \cos^2 \frac{	heta}{2} + i 2 \sin \frac{	heta}{2} \cos \frac{	heta}{2}} = \frac{\cos \frac{	heta}{2} - i \sin \frac{	heta}{2}}{\cos \frac{	heta}{2} + i \sin \frac{	heta}{2}} = \frac{e^{-i 	heta/2}}{e^{i 	heta/2}} = e^{-i 	heta}$.Then $z_2^{16} = (e^{-i 	heta})^{16} = e^{-i 16 	heta} = \cos 16 	heta - i \sin 16 	heta$.Adding the two terms: $z_1^8 + z_2^{16} = (\cos 16 	heta + i \sin 16 	heta) + (\cos 16 	heta - i \sin 16 	heta) = 2 \cos 16 	heta$.

List-$I$	List-$II$
$(P)$ $\|z\|^2$ is equal to	$(1)$ $12$
$(Q)$ $\|z-\bar{z}\|^2$ is equal to	$(2)$ $4$
$(R)$ $\|z\|^2+\|z+\bar{z}\|^2$ is equal to	$(3)$ $8$
$(S)$ $\|z+1\|^2$ is equal to	$(4)$ $10$
	$(5)$ $7$

$\left(\frac{\cos \theta+i \sin \theta}{\sin \theta+i \cos \theta}\right)^8+\left(\frac{1+\cos \theta-i \sin \theta}{1+\cos \theta+i \sin \theta}\right)^{16}=$

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