Question

Let a,b $(a\ne b)$ are two non-zero complex numbers satisfying $|a^2-b^2|=|a^2+b^2-2ab|$ then

• A. $\frac{a}{b}$ is purely real
• B. $\frac{a}{b}$ is purely imaginary
• C. $|arg\left(a\right)-arg\left(b\right)|=\pi$
• D. $|arg\left(a\right)-arg\left(b\right)|=\frac{\pi }{4}$

Answer 1

Answer: (D)

$|arg\left(a\right)-arg\left(b\right)|=\frac{\pi }{4}$

Given: $|a^2-b^2|=|a^2+b^2-2ab|$

Now, we form the equation as

$⟹|a^2-b^2|x^2-|a^2+b^2-2ab|x=0$

$⟹x(|a^2-b^2|x-|a^2+b^2-2ab|)=0$

$⟹x=0$ or $x=\frac{|a^2+b^2-2ab|}{|a^2-b^2|}$

$⟹\frac{|a^2+b^2-2ab|}{|a^2-b^2|}=0$

$⟹\frac{|(a-b{)}^2|}{|(a-b)(a+b)|}=0$

$⟹\frac{|(a-b{)}^2|}{|a-b||a+b|}=0$

$⟹\frac{|a-b|}{|a+b|}=0$

$⟹|a-b|=0$

$⟹|a|=|b|$

$⟹|\frac{a}{b}|=1$

Hence, $⟹arg|\frac{a}{b}|=|arg\left(a\right)-arg\left(b\right)|=\frac{\pi }{4}$