If z = cos theta + i sin theta,then z^r + (ba | vedclass

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(B) Given $z = \cos 	heta + i \sin 	heta$. By De Moivre's Theorem,$z^r = (\cos 	heta + i \sin 	heta)^r = \cos(r 	heta) + i \sin(r 	heta)$. The c

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$ 2 \cos r 	heta $

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(B) Given $z = \cos 	heta + i \sin 	heta$. By De Moivre's Theorem,$z^r = (\cos 	heta + i \sin 	heta)^r = \cos(r 	heta) + i \sin(r 	heta)$. The conjugate is $\bar{z} = \cos 	heta - i \sin 	heta$. Thus,$(\bar{z})^r = (\cos 	heta - i \sin 	heta)^r = \cos(r 	heta) - i \sin(r 	heta)$. Adding these two expressions: $z^r + (\bar{z})^r = (\cos(r 	heta) + i \sin(r 	heta)) + (\cos(r 	heta) - i \sin(r 	heta)) = 2 \cos(r 	heta)$.

If $z = \cos \theta + i \sin \theta$,then $z^r + (\bar{z})^r = $

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