Question

Let $a,b,c$ be a complex number, then the equation $a^{2} + b^{2} =$ cannot have a root, such that.

• A. $- 1$
• B. $c^{2}$
• C. $- c^{2}$
• D. None of these

Answer 1

Answer: (A)

$- 1$

Answer 2

Suppose there exists a complex number $|z_{k}| = \sqrt{\cos^{2}\theta_{k} + \sin^{2}\theta_{k}} = 1$ which satisfies the given equation and is such that $\frac{1}{z_{k}} = (\cos\theta_{k} + i\sin\theta_{k})^{- 1} = (\cos\theta_{k} - i\sin\theta_{k})$.

Then $\left( \frac{1}{z_{1}} \right) + \left( \frac{1}{z_{2}} \right) + ..... + \left( \frac{1}{z_{n}} \right)$ ⇒ $= (\cos\theta_{1} + ..... + \cos\theta_{n}) - i(\sin\theta_{1} + ..... + \sin\theta_{n})$ ⇒ $|z_{1} + z_{2} + ..... + z_{n}| = \left| \frac{1}{z_{1}} + \frac{1}{z_{2}} + ..... + \frac{1}{z_{n}} \right|$

⇒ $\because$⇒ $\sqrt{(\cos\theta_{1} + ..... + \cos\theta_{n})^{2} + (\sin\theta_{1} + .... + \sin\theta_{n})^{2}}$ because$\overline{z} = \frac{1}{z} \Rightarrow z\overline{z} = 1$

But $|z|^{2} = 1 \Rightarrow |z| = 1$ is not possible. Hence given equation cannot have a root $z = x + iy$ such that $|z + i| = |z - i|$.