For any two complex numbers `z_1,z_2` and any real numbers `aa n db ,|a z_1-b z_2|^2+|b z_1+a z_2|^2=`

Options

A. `(a^(2)+b^(2))|z_(1)z_(2)|`
B. `(a^(2)+b^(2))|z_(1)^(2)+z_(2)^(2)|`
C. `(a^(2)+b^(2))(|z_(1)|^(2)+|z_(2)|^(2))`
D. `2ab|z_(1)z_(2)|`

Answer

C

Explanation

The given function is periodic, with period `2pi`. So the difference between the greatest and least values of the function is the difference between these values on the interval `[0,2pi]` We have
`f'(x)=-(sinx+sin2x-sin3x)=-4sinxsin(3x//2)sin(x//2)`
Hence x = 0 , `2pi//3,pi` and `2pi` are the critical points.
Also, `f(0)=1+1//2-1//3=7//6,f(2pi//3)=-13//12,f(pi)=-1//6`and `f(2pi)=7//6`.
Hence the greatest value is `7//6` and the least value is `-13//12`. Thus the difference is `(7)/(6)-(-(13)/(12))=(27)/(12)=(9)/(4)`