A) \[2\sqrt{3}-2i\]
B) \[2\sqrt{3}+2i\]
C) \[-2\sqrt{3}+2i\]
D) \[-\sqrt{3}+i\]
Correct Answer: C
Solution :
\[|z|=4\]and \[arg\,z=\frac{5\pi }{6}={{150}^{o}}\] Let \[z=x+iy\], then \[|z|=r=\sqrt{{{x}^{2}}+{{y}^{2}}}=4\] and \[\theta =\frac{5\pi }{6}={{150}^{o}}\] \[\therefore \] \[x=r\cos \theta =4\cos \,\,{{150}^{o}}=-2\sqrt{3}\]. and \[y=r\sin \theta =4\]\[\sin {{150}^{o}}=4\frac{1}{2}=2\] \[\therefore \] \[z=x+iy=-2\sqrt{3}+2i\] Trick: Since\[arg\,z=\frac{5\pi }{6}={{150}^{o}}\], here the complex number must lie in second quadrant, so (a) and (b) rejected. Also \[|z|=4\] which satisfies (c) only.
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