A) \[6\]
B) \[\sqrt{2}\]
C) \[\sqrt{6}\]
D) \[\sqrt{3}\]
E) \[\sqrt{2}+\sqrt{3}\]
Correct Answer: C
Solution :
Given that \[{{z}_{1}}=\sqrt{2}\left( \cos \frac{\pi }{4}+i\sin \frac{\pi }{4} \right)\] \[=1+i\]\[\therefore \]\[|{{z}_{1}}|=\sqrt{{{1}^{2}}+{{1}^{2}}}=\sqrt{2}\] and \[{{z}_{2}}=\sqrt{3}\left( \cos \frac{\pi }{3}+i\sin \frac{\pi }{3} \right)\] \[=\frac{\sqrt{3}}{2}+\frac{3i}{2}\] \[\therefore \] \[|{{z}_{2}}|=\sqrt{\left( \frac{\sqrt{3}}{2} \right)+{{\left( \frac{3}{2} \right)}^{2}}}\] \[=\sqrt{\frac{3}{4}+\frac{9}{4}}=\frac{2\sqrt{3}}{2}=\sqrt{3}\] \[\therefore \] \[|{{z}_{1}}{{z}_{2}}|=|{{z}_{1}}|.|{{z}_{2}}|\] \[=\sqrt{2}.\sqrt{3}=\sqrt{6}\]
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