Complex Numbers And Quadratic Equations question 67
Question: If $ |z_1|=|z_2|=……….=|z_{n}|=1, $ then the value of $ |z_1+z_2+z_3+………….+z_{n}| $ =
Options:
A) 1
B) $ |z_1|+|z_2|+…….+|z_{n}| $
C) $ | \frac{1}{z_1}+\frac{1}{z_2}+………+\frac{1}{z_{n}} | $
D) None of these
Show Answer
Answer:
Correct Answer: C
Solution:
We have $ |z_{k}|=1,k=1,2,….n $
Þ $ |z_{k}{{|}^{2}}=1\Rightarrow z_{k}{{\overline{z}}{k}}=1\Rightarrow {{\overline{z}}{k}}=\frac{1}{z_{k}} $ Therefore $ |z_1+z_2+….+z_{n}|=|\overline{z_1+z_2+….+z_{n}}| $ $ (\because |z|=|\overline{z}|) $ $ =|{{\overline{z}}1}+\overline{z_2}+…..+{{\overline{z}}{n}}|=| \frac{1}{z_1}+\frac{1}{z_2}+….+\frac{1}{z_{n}} | $ Aliter: Let $ z_{k}=\cos {\theta_{k}}+i\sin {\theta_{k}},k=1,2,….n $ So that $ |z_{k}|=\sqrt{{{\cos }^{2}}{\theta_{k}}+{{\sin }^{2}}{\theta_{k}}}=1 $ Then $ \frac{1}{z_{k}}={{(\cos {\theta_{k}}+i\sin {\theta_{k}})}^{-1}}=(\cos {\theta_{k}}-i\sin {\theta_{k}}) $ Now, $ z_1+z_2+…..+z_{n} $ $ =(\cos {\theta_1}+…..+\cos {\theta_{n}})-i(\sin {\theta_1}+…..+\sin {\theta_{n}}) $ and $ ( \frac{1}{z_1} )+( \frac{1}{z_2} )+…..+( \frac{1}{z_{n}} ) $ $ =(\cos {\theta_1}+…..+\cos {\theta_{n}})-i(\sin {\theta_1}+…..+\sin {\theta_{n}}) $ Hence $ |z_1+z_2+…..+z_{n}|=| \frac{1}{z_1}+\frac{1}{z_2}+…..+\frac{1}{z_{n}} | $ Since each side is equal to $ \sqrt{{{(\cos {\theta_1}+…..+\cos {\theta_{n}})}^{2}}+{{(\sin {\theta_1}+….+\sin {\theta_{n}})}^{2}}} $
BETA
