Complex Numbers And Quadratic Equations question 63
Question: If $ z_1\text{ and }z_2 $ be complex numbers such that $ z_1\ne z_2 $ and $ |z_1|=|z_2| $ . If $ z_1 $ has positive real part and $ z_2 $ has negative imaginary part, then $ \frac{(z_1+z_2)}{(z_1-z_2)} $ may be [IIT 1986]
Options:
A) Purely imaginary
B) Real and positive
C) Real and negative
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
Let $ z_1=a+ib=(a,b) $ and $ z_2=c-id=(c,-d) $ Where $ a>0 $ and $ d>0 $ ……(i) Then $ |z_1|=|z_2| $
Þ $ a^{2}+b^{2}=c^{2}+d^{2} $ Now $ \frac{z_1+z_2}{z_1-z_2}=\frac{(a+ib)+(c-id)}{(a+ib)-(c-id)} $ $ =\frac{[(a+c)+i(b-d)][(a-c)-i(b+d)]}{[(a-c)+i(b+d)][(a-c)-i(b+d)]} $ $ =\frac{(a^{2}+b^{2})-(c^{2}+d^{2})-2(ad+bc)i}{a^{2}+c^{2}-2ac+b^{2}+d^{2}+2bd} $ $ \frac{-(ad+bc)i}{a^{2}+b^{2}-ac+bd} $ [using (i)]
$ \therefore $ $ \frac{(z_1+z_2)}{(z_1-z_2)} $ is purely imaginary. However if $ ad+bc=0 $ , then $ \frac{(z_1+z_2)}{(z_1-z_2)} $ will be equal to zero. According to the conditions of the equation, we can have $ ad+bc=0 $ Trick: Assume any two complex numbers satisfying both conditions i.e., $ z_1\ne z_2 $ and $ |z_1|=|z_2| $ Let $ z_1=2+i,z_2=1-2i, $
$ \therefore \frac{z_1+z_2}{z_1-z_2}=\frac{3-i}{1+3i}=-i $ Hence the result.
BETA
