Convert the complex number z = frac{i-1}{cos | vedclass

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(B) Given $z = \frac{i-1}{\cos \frac{\pi}{3} + i \sin \frac{\pi}{3}}$.Since $\cos \frac{\pi}{3} = \frac{1}{2}$ and $\sin \frac{\pi}{3} = \frac{\sqrt{3

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$\sqrt{2} \left( \cos \frac{5 \pi}{12} + i \sin \frac{5 \pi}{12} \right)$

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(B) Given $z = \frac{i-1}{\cos \frac{\pi}{3} + i \sin \frac{\pi}{3}}$.Since $\cos \frac{\pi}{3} = \frac{1}{2}$ and $\sin \frac{\pi}{3} = \frac{\sqrt{3}}{2}$,we have $z = \frac{i-1}{\frac{1}{2} + i \frac{\sqrt{3}}{2}} = \frac{2(i-1)}{1 + i \sqrt{3}}$.Multiplying numerator and denominator by the conjugate $(1 - i \sqrt{3})$:$z = \frac{2(i-1)(1 - i \sqrt{3})}{(1 + i \sqrt{3})(1 - i \sqrt{3})} = \frac{2(i + \sqrt{3} - 1 - i^2 \sqrt{3})}{1 + 3} = \frac{2(\sqrt{3}-1 + i(\sqrt{3}+1))}{4} = \frac{\sqrt{3}-1}{2} + i \frac{\sqrt{3}+1}{2}$.For polar form $z = r(\cos 	heta + i \sin 	heta)$,$r = |z| = \sqrt{(\frac{\sqrt{3}-1}{2})^2 + (\frac{\sqrt{3}+1}{2})^2} = \sqrt{\frac{3-2\sqrt{3}+1 + 3+2\sqrt{3}+1}{4}} = \sqrt{\frac{8}{4}} = \sqrt{2}$.Now,$\cos 	heta = \frac{\sqrt{3}-1}{2\sqrt{2}}$ and $\sin 	heta = \frac{\sqrt{3}+1}{2\sqrt{2}}$.Using $\cos(A+B) = \cos A \cos B - \sin A \sin B$ and $\sin(A+B) = \sin A \cos B + \cos A \sin B$,we find $	heta = \frac{5\pi}{12}$.Thus,$z = \sqrt{2} (\cos \frac{5\pi}{12} + i \sin \frac{5\pi}{12})$.

Convert the complex number $z = \frac{i-1}{\cos \frac{\pi}{3} + i \sin \frac{\pi}{3}}$ into polar form.

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