Complex Numbers And Quadratic Equations question 430
Question: If $ \frac{{{(p+i)}^{2}}}{2p-i}=\mu +i\lambda , $ then $ {{\mu }^{2}}+{{\lambda }^{2}} $ is equal to
Options:
A) $ \frac{{{(p^{2}+1)}^{2}}}{4p^{2}-1} $
B) $ \frac{{{(p^{2}-1)}^{2}}}{4p^{2}-1} $
C) $ \frac{{{(p^{2}-1)}^{2}}}{4p^{2}+1} $
D) $ \frac{{{(p^{2}+1)}^{2}}}{4p^{2}+1} $
Show Answer
Answer:
Correct Answer: D
Solution:
$ \mu +i\lambda =\frac{{{(p+i)}^{2}}}{2p-i}=\frac{(p^{2}-1+2pi)(2p+i)}{(2p-i)(2p+i)} $ $ =\frac{2p(p^{2}-2)+i(5p^{2}-1)}{4p^{2}+1} $ \ $ {{\mu }^{2}}+{{\lambda }^{2}}=\frac{4p^{2}{{(p^{2}-2)}^{2}}+{{(5p^{2}-1)}^{2}}}{{{(4p^{2}+1)}^{2}}} $ $ =\frac{4p^{6}+6p^{2}+9p^{4}+1}{{{(4p^{2}+1)}^{2}}} $ $ =\frac{p^{4}(4p^{2}+1)+2p^{2}(4p^{2}+1)+(4p^{2}+1)}{{{(4p^{2}+1)}^{2}}} $ $ =\frac{p^{4}+2p^{2}+1}{4p^{2}+1}=\frac{{{(p^{2}+1)}^{2}}}{4p^{2}+1} $ .
BETA
