Complex Numbers And Quadratic Equations question 62
Question: The solution of the equation $ |z|-z=1+2i $ is [MP PET 1993]
Options:
A) $ 2-\frac{3}{2}i $
B) $ \frac{3}{2}+2i $
C) $ \frac{3}{2}-2i $
D) $ -2+\frac{3}{2}i $
Show Answer
Answer:
Correct Answer: C
Solution:
$ |z|-z=1+2i $ Let $ z=x+iy $ , therefore $ |x+iy|-(x+iy)=1+2i $ Equating real and imaginary parts, we get $ \sqrt{x^{2}+y^{2}}-x=1 $ and $ y=-2 $ Þ $ x=\frac{3}{2} $ Hence complex number $ z=\frac{3}{2}-2i $ . Trick: Since $ | \frac{3}{2}-2i |-( \frac{3}{2}-2i ) $ $ =\sqrt{\frac{9}{4}+4}-\frac{3}{2}+2i=\frac{5}{2}-\frac{3}{2}+2i=1+2i $
BETA
