First slide

Algebra of complex numbers

Question

If z₁ and z₂ are complex numbers such that $z_{1} \neq z_{2}$ and |z₁| = |z₂|. If z₁ has positive real part and z₂ has negative imaginary part, then $\frac{z_{1} + z_{2}}{z_{1} - z_{2}}$ may be

Moderate

A

purely imaginary
B

real and positive
C

real and negative
D

none of these

Solution

Let $z_{1} = a + ib and z_{2} = c - id$ , where a > 0 and d > 0. Then,
$|z_{1}| = |z_{2}| \Rightarrow a^{2} + b^{2} = c^{2} + d^{2}$ (1)
$\begin{matrix} Now, \frac{z_{1} + z_{2}}{z_{1} - z_{2}} = \frac{(a + ib) + (c - id)}{(a + ib) - (c - id)} \\ = \frac{[(a + c) + i (b - d)] [(a - c) - i (b + d)]}{[(a - c) + i (b + d)] [(a - c) - i (b + d)]} \\ = \frac{(a^{2} + b^{2}) - (c^{2} + d^{2}) - 2 (ad + bc) i}{a^{2} + c^{2} - 2 ac + b^{2} + d^{2} + 2 bd} \\ = \frac{- (ad + bc) i}{a^{2} + b^{2} - ac + bd} [Using (1)] \end{matrix}$
Hence, $\frac{z_{1} + z_{2}}{z_{1} - z_{2}}$ is purely imaginary. However, if ad + bc = 0, then $\frac{z_{1} + z_{2}}{z_{1} - z_{2}}$ will be equal to zero. According to the conditions of the equation, we can have ad + bc = 0.

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