By given data,

$b+ic=(1+a)z$

R.H.S=

$\dfrac {1+iz}{1-iz} = \dfrac {1+i \left \{ \frac {b+ic}{1+a} \right \}}{1-i \left \{ \frac {b+ic}{1+a} \right \}} $

$\dfrac {1+iz}{1-iz} = \dfrac {1+a+ib-c}{1+a-ib+c} = \dfrac {(1+a-c)+ib}{(1+a+c)-ib} \cdot \dfrac {(1+a+c)+ib}{(1+a+c)+ib}$

$ =\dfrac {1+a^2 - b^2 + 2a + 2ib + 2aib - c^2}{1+a^2 +c^2 + 2a + 2c+2ac +b^2}$

Since by data, $a^2+b^2+c^2=1$ in the numerator we put $a^2=1-b^2-c^2$ And in the denominator we put $a^2 + b^2 + +c^2=1$

$\dfrac {1+iz}{1-iz} = \dfrac {2a^{2}+2a + 2ib + 2aib}{2+2a+2c+2ac} = \dfrac {a (a+1)+ib (1+a)}{1(1+a)+c(1+a)} = \dfrac {(a+1)(a+ib)}{(1+a) (1+c)}$

$ =\dfrac {a+ib}{1+c}= \text{L.H.S.}$

$ Hence, \dfrac {a+ib}{1+c} = \dfrac {1+iz}{1-iz} $