Question

If $a,b,c$ are real, $a^2+b^2+c^2=1$ and $b+ic=(1+a)z$, then $\frac{1+zi}{1-zi}=$

• A. $\frac{a-ib}{1+c}$
• B. $\frac{a+ib}{1+c}$
• C. $\frac{a+ib}{1-c}$
• D. $\frac{a-ib}{1-c}$

Answer 1

The correct option is B $\frac{a+ib}{1+c}$

We have, $\frac{1+zi}{1-zi}=\frac{1+i\frac{b+ic}{1+a}}{1-i\frac{b+ic}{1+a}}=\frac{1+a-c+ib}{1+a+c-ib}$

$=\frac{[1+a-c+ib][1+a+c+ib]}{[1+a+c-ib][1+a+c+ib]}=\frac{z_0}{{(1+a+c)}^2+b^2}$ ...(1)

Now, $Real\left(z_0\right)=(1+a-c)(1+a+c)-b^2$

$={(1+a)}^2-c^2-b^2=1+a^2+2a-{(1-a)}^2=2a^2+2a=2a(1+a)$

and, $Im\left(z_0\right)=(1+a-c)b+(1+a+c)b$

$=2b(1+a)$.

Thus $z_0=2a(1+a)+i2b(1+a)=2(1+a)(a+ib)$

Also, denominator of (1) $=1+a^2+c^2+2a+2c+2ac+b^2$

$=2+2a+2c+2ac=2(1+a)(1+c)$

Therefore, $\frac{1+zi}{1-zi}=\frac{2(1+a)(a+ib)}{2(1+a)(1+c)}=\frac{a+ib}{1+c}$