# Solve this

Question:

If $(a+i b)=\frac{c+i}{c-i}$, where $c$ is real, prove that $a^{2}+b^{2}=1$ and $\frac{b}{a}=\frac{2 c}{c^{2}-1}$

Solution:

Consider the given equation,

$a+i b=\frac{c+i}{c-i}$

Now, rationalizing

$a+i b=\frac{c+i}{c-i} \times \frac{c+i}{c+i}$

$=\frac{(c+i)(c+i)}{(c-i)(c+i)}$

$=\frac{(c+i)^{2}}{(c)^{2}-(i)^{2}}$

$\left[(a-b)(a+b)=a^{2}-b^{2}\right]$

$=\frac{c^{2}+2 i c+i^{2}}{c^{2}-i^{2}}$

$a+i b=\frac{c^{2}+2 i c+(-1)}{c^{2}-(-1)} \quad\left[i^{2}=-1\right]$

$a+i b=\frac{c^{2}+2 i c-1}{c^{2}+1}$

$a+i b=\frac{\left(c^{2}-1\right)}{c^{2}+1}+i \frac{2 c}{c^{2}+1}$

On comparing both the sides, we get

$a=\frac{\left(c^{2}-1\right)}{c^{2}+1} \& b=\frac{2 c}{c^{2}+1}$

Now, we have to prove that $a^{2}+b^{2}=1$

Taking LHS,

$a^{2}+b^{2}$

Putting the value of a and b, we get

$\left[\frac{\left(c^{2}-1\right)}{c^{2}+1}\right]^{2}+\left[\frac{2 c}{c^{2}+1}\right]^{2}$

$=\frac{1}{\left(c^{2}+1\right)^{2}}\left[\left(c^{2}-1\right)^{2}+(2 c)^{2}\right]$

$=\frac{1}{\left(c^{2}+1\right)^{2}}\left[c^{4}+1-2 c^{2}+4 c^{2}\right]$

$=\frac{1}{\left(c^{2}+1\right)^{2}}\left[c^{4}+1+2 c^{2}\right]$

$=\frac{1}{\left(c^{2}+1\right)^{2}}\left[\left(c^{2}+1\right)^{2}\right]$

= 1

= RHS

Now, we have to prove $\frac{b}{a}=\frac{2 c}{c^{2}-1}$

Taking LHS, $\frac{b}{a}$

Putting the value of a and b, we get

$\frac{b}{a}=\frac{\frac{2 c}{c^{2}+1}}{\frac{\left(c^{2}-1\right)}{c^{2}+1}}=\frac{2 c}{c^{2}+1} \times \frac{c^{2}+1}{c^{2}-1}=\frac{2 c}{c^{2}-1}=R H S$

Hence Proved