Complex Numbers And Quadratic Equations question 658
Question: Let $ x_1 $ and $ y_1 $ be real numbers. If $ z_1 $ and $ z_2 $ are complex numbers such that $ |z_1|=|z_2|=4, $ then $ |x_1z_1-y_1z_2{{|}^{2}}+|y_1z_1+x_1z_2{{|}^{2}}= $
Options:
A) $ 32(x_1^{2}+y_1^{2}) $
B) $ 16(x_1^{2}+y_1^{2}) $
C) $ 4(x_1^{2}+y_1^{2}) $
D) $ 32(x_1^{2}+y_1^{2})|z_1+z_2{{|}^{2}} $
Show Answer
Answer:
Correct Answer: A
Solution:
$ {{| x_1y_1-y_1z_2 |}^{2}}+{{| y_1z_1-x_1z_2 |}^{2}} $ $ ={{| x_1y_1 |}^{2}}+{{| y_1z_2 |}^{2}}-2Re(x_1y_1z_1z_2) $ $ +|y_1z_1{{|}^{2}}+{{| x_1z_2 |}^{2}}+2Re(x_1y_1z_1z_2) $ $ =x_2^{1}{{| z_1 |}^{2}}+y_1^{2}{{| z_2 |}^{2}}+y_1^{2}{{| z_1 |}^{2}}+x_1^{2}{{| z_2 |}^{2}} $ $ =x_2^{1}{{| z_1 |}^{2}}+y_1^{2}{{| z_2 |}^{2}}+y_1^{2}{{| z_1 |}^{2}}+x_1^{2}{{| z_2 |}^{2}} $ $ =2( x_1^{2}+y_1^{2} )( 4^{2} )=32( x_1^{2}+y_1^{2} ) $
BETA
