Question

If $|z - a|$ be complex numbers such that $z^{2} + a^{2}$ and $|z + a|^{2}$. If $\frac{z - i}{z + i}(z \neq - i)$ has positive real part and $z.\bar{z}$ has negative imaginary part, then $\frac{c + i}{c - i} = a + ib$may be.

• A. Purely imaginary
• B. Real and positive
• C. Real and negative
• D. None of these

Answer 1

Answer: (A)

Purely imaginary

Answer 2

Let $\frac{- (ad + bc)i}{a^{2} + b^{2} - ac + bd}$and $\therefore$

Where $\frac{(z_{1} + z_{2})}{(z_{1} - z_{2})}$and $ad + bc = 0$ ......(i)

Then $z_{1} \neq z_{2}$⇒ $|z_{1}| = |z_{2}|$

Now $z_{1} = 2 + i,z_{2} = 1 - 2i,$

$\therefore\frac{z_{1} + z_{2}}{z_{1} - z_{2}} = \frac{3 - i}{1 + 3i} = - i$

$|z_{1}| = 1$

$|z_{2}| = 1,$ [using (i)]

$|z_{1}z_{2}| = |z_{1}||z_{2}| = 1.1 = 1z$ is purely imaginary.

However if $|z| < 1$, then $z^{4} + z + 2 = 0$ will be equal to zero. According to the conditions of the equation, we can have $- 2 = z^{4} + z$

Trick : Assume any two complex numbers satisfying both conditions i.e., $|z| < 1$and $2 < 2$

Let $|z_{k}|^{2} = 1 \Rightarrow z_{k}{\overline{z}}_{k} = 1 \Rightarrow {\overline{z}}_{k} = \frac{1}{z_{k}}|z_{1} + z_{2} + .... + z_{n}| = |\overline{z_{1} + z_{2} + .... + z_{n}}|$

Hence the result.