First slide

Algebra of complex numbers

Question

If $a^{2} + b^{2} = 1,$ then $\frac{1 + b + ia}{1 + b - ia}$

Moderate

A

1
B

2
C

b + ia
D

a + ib

Solution

Given that a² + b² = 1. Therefore,
$\begin{matrix} \frac{1 + b + ia}{1 + b - ia} = \frac{(1 + b + ia) (1 + b + ia)}{(1 + b - ia) (1 + b + ia)} \\ = \frac{(1 + b)^{2} - a^{2} + 2 ia (1 + b)}{1 + b^{2} + 2 b + a^{2}} \\ = \frac{(1 - a^{2}) + 2 b + b^{2} + 2 ia (1 + b)}{2 (1 + b)} \\ = \frac{2 b^{2} + 2 b + 2 ia (1 + b)}{2 (1 + b)} = b + ia \end{matrix}$

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