A) \[\cos 2\theta \]
B) \[2\cos 2\theta \]
C) \[2\cos \theta \]
D) \[2\sin \theta \]
E) \[2\sin 2\theta \]
Correct Answer: B
Solution :
We have\[z=r(\cos \theta +i\sin \theta )\] \[\therefore \]\[\overline{z}=r(\cos \theta -i\sin \theta )\] \[\therefore \] \[\frac{z}{z}+\frac{\overline{z}}{z}\] \[=\frac{r(\cos \theta +i\sin \theta )}{r(\cos \theta -i\sin \theta )}+\frac{r(\cos \theta -i\sin \theta )}{r(\cos \theta +i\sin \theta )}\] \[=\frac{{{(\cos \theta +i\sin \theta )}^{2}}+(\cos \theta -i\sin {{\theta }^{2}})}{{{\cos }^{2}}\theta +{{\sin }^{2}}\theta }\] \[={{\cos }^{2}}\theta -{{\sin }^{2}}\theta +2i\cos \theta \sin \theta +{{\cos }^{2}}\theta \] \[-{{\sin }^{2}}\theta +2i\cos \theta \sin \theta \] \[=2({{\cos }^{2}}\theta -{{\sin }^{2}}\theta )\] \[=2\cos 2\theta \]
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