Question

$z_1$ and $z_2$ are the roots of the equation $z^2-az+b=0$, where $\left|z_1\right|=\left|z_2\right|=1$ a, b are nonzero complex numbers, then

• A. $\left|a\right|\le 1$
• B. $\left|a\right|\le 2$
• C. arg$\left(a^2\right)=arg\left(b\right)$
• D. arg a = arg $\left(b^2\right)$

Answer 1

Answer: (B), (C)

The correct options are
B $\left|a\right|\le 2$
C arg$\left(a^2\right)=arg\left(b\right)$

$z^2-az+b=0$ ...(1)
$z_1\&z_2$ where $\left|z_1\right|=\left|z_2\right|=1$
Sum of roots of the equation (1)=$z_1+z_2=a$
Products of the roots of the equation (1)=$z_1{z}_2=b$
$|z_1+z_2|\le \left|z_1\right|+\left|z_2\right|\Rightarrow \left|a\right|\le 1+1\Rightarrow \left|a\right|\le 2$
Let $z_1=\cos A+i\sin A\&z_2=\cos B+i\sin B$
$arg\left(\frac{a}{b}\right)=arg\left(\frac{z_1+z_2}{z_1{z}_2}\right)=arg\left(\frac{\cos A+i\sin A+\cos B+i\sin B}{(\cos A+i\sin A)(\cos B+i\sin B)}\right)$
$\Rightarrow arg\left(\frac{a}{b}\right)=arg\left(\frac{\cos A+\cos B+i\sin A+i\sin B}{\cos (A+B)+i\sin (A+B)}\right)$
$\Rightarrow arg\left(\frac{a}{b}\right)=\frac{\arctan \left(\frac{\sin A+\sin B}{\cos A+\cos B}\right)}{\arctan \left(\frac{\sin (A+B)}{\cos (A+B)}\right)}=\frac{\arctan \left(\frac{\sin A+\sin B}{\cos A+\cos B}\right)}{A+B}$
$\Rightarrow arg\left(\frac{a}{b}\right)=\frac{\arctan \left(\frac{2\sin \left(\frac{A+B}{2}\right)\cos \left(\frac{A-B}{2}\right)}{2\cos \left(\frac{A+B}{2}\right)\cos \left(\frac{A-B}{2}\right)}\right)}{A+B}=\frac{\arctan \left(\tan \left(\frac{A+B}{2}\right)\right)}{A+B}$
$\Rightarrow arg\left(\frac{a}{b}\right)=\frac{A+B}{2(A+B)}=\frac{1}{2}$
$\therefore arg\left(a^2\right)=argb$
Hence, options 'B' and 'C' are correct.