If z(1+a)=b+ic and a2+b2+c2=1 then (1+iz)(1−iz)=

# If  then $\frac{\left(1+\mathrm{iz}\right)}{\left(1-\mathrm{iz}\right)}=$

1. A

$\frac{\mathrm{a}+\mathrm{ib}}{1+\mathrm{c}}$

2. B

$\frac{\mathrm{b}-\mathrm{ic}}{1+\mathrm{a}}$

3. C

$\frac{\mathrm{a}+\mathrm{ic}}{1+\mathrm{b}}$

4. D

none of these

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### Solution:

$\begin{array}{l}\frac{1+\mathrm{iz}}{1-\mathrm{iz}}=\frac{1+\mathrm{i}\left(\mathrm{b}+\mathrm{ic}\right)/\left(1+\mathrm{a}\right)}{1-\mathrm{i}\left(\mathrm{b}+\mathrm{ic}\right)/\left(1+\mathrm{a}\right)}\\ =\frac{1+\mathrm{a}-\mathrm{c}+\mathrm{ib}}{1+\mathrm{a}+\mathrm{c}-\mathrm{ib}}\\ =\frac{\left(1+\mathrm{a}-\mathrm{c}+\mathrm{ib}\right)\left(1+\mathrm{a}+\mathrm{c}+\mathrm{ib}\right)}{\left(1+\mathrm{a}+\mathrm{c}{\right)}^{2}+{\mathrm{b}}^{2}}\end{array}$

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