If 2+2 sqrt{3} i=k(cos theta+i sin theta) whe | vedclass

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(B) Given,$k(\cos 	heta+i \sin 	heta)=2+2 \sqrt{3} i$. Comparing the modulus and argument,we have $k = \sqrt{2^2 + (2\sqrt{3})^2} = \sqrt{4 + 12} =

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$\frac{1}{\sqrt{3}}$

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(B) Given,$k(\cos 	heta+i \sin 	heta)=2+2 \sqrt{3} i$. Comparing the modulus and argument,we have $k = \sqrt{2^2 + (2\sqrt{3})^2} = \sqrt{4 + 12} = \sqrt{16} = 4$. Also,$\cos 	heta = \frac{2}{4} = \frac{1}{2}$ and $\sin 	heta = \frac{2\sqrt{3}}{4} = \frac{\sqrt{3}}{2}$,which implies $	heta = \frac{\pi}{3}$. Now,we need to evaluate $\frac{1}{\sqrt{3}}[\cos 6 	heta+i \sin 6 	heta]$. Substituting $	heta = \frac{\pi}{3}$,we get $6	heta = 6(\frac{\pi}{3}) = 2\pi$. Thus,$\frac{1}{\sqrt{3}}[\cos 2\pi + i \sin 2\pi] = \frac{1}{\sqrt{3}}[1 + i(0)] = \frac{1}{\sqrt{3}}$.

If $2+2 \sqrt{3} i=k(\cos \theta+i \sin \theta)$ where $k > 0$,then find the value of $\frac{1}{\sqrt{3}}[\cos 6 \theta+i \sin 6 \theta]$.

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