Question

If $z(1 + a) = b + ic$ and ${a^2} + {b^2} + {c^2} = 1$, then $\dfrac{{1 + iz}}{{1 - iz}} =$
A) $\dfrac{{a + ib}}{{1 + c}}$
B) $\dfrac{{b - ic}}{{1 + a}}$
C) $\dfrac{{a + ic}}{{1 + b}}$
D) None of these

Answer 1

The given question is related to complex numbers. First we must find the value of z from the given equation. Later we will substitute the z value which we previously got into $\dfrac{{1 + iz}}{{1 - iz}}$ to get the desired answer. Finally we will select the option which matches with our answer.
Complex numbers are the combination of both real numbers and imaginary numbers. The complex number is of the standard form: $a + ib$
Where
$a$ and $b$ are real numbers.
$i$ is an imaginary unit.

Complete step-by-step answer:
It is given that; $z(1 + a) = b + ic$ and ${a^2} + {b^2} + {c^2} = 1$
Consider the given relation $z(1 + a) = b + ic$,
$z(1 + a) = b + ic$
Solving this relation for z,
Hence divided the relation by $(1 + a)$ on the both sides, we get,
$\Rightarrow \dfrac{{z(1 + a)}}{{(1 + a)}} = \dfrac{{b + ic}}{{(1 + a)}}$
Cancelling the similar terms we get,
$\Rightarrow z = \dfrac{{b + ic}}{{(1 + a)}}$
We have to find the value of $\dfrac{{1 + iz}}{{1 - iz}}$.
Now,
$\dfrac{{1 + iz}}{{1 - iz}}$
Substitute the value of z we get,
$\Rightarrow \dfrac{{1 + i\dfrac{{b + ic}}{{1 + a}}}}{{1 - i\dfrac{{b + ic}}{{1 + a}}}}$
Simplifying we get,
$\Rightarrow \dfrac{{1 + a - c + ib}}{{1 + a + c - ib}}$
We will multiply $(1 + a + c) + ib$ with both the numerator and denominator we get,
$\Rightarrow \dfrac{{(1 + a - c + ib)(1 + a + c + ib)}}{{(1 + a + c - ib)(1 + a + c + ib)}}$
Simplifying we get,
$\Rightarrow \dfrac{{1 + 2a + {a^2} - {b^2} - {c^2} + 2ib + 2iab}}{{1 + {a^2} + {c^2} + {b^2} + 2ac + 2(a + c)}}$
We know that, ${a^2} + {b^2} + {c^2} = 1$
Substitute this value in the above step we get,
$\Rightarrow \dfrac{{2a + 2{a^2} + 2ib + 2iab}}{{2 + 2ac + 2(a + c)}}$
Eliminating 2 from both the numerator and denominator we get,
$\Rightarrow \dfrac{{a + {a^2} + ib + iab}}{{1 + ac + (a + c)}}$
Simplifying we get,
$\Rightarrow \dfrac{{a(1 + a) + ib(1 + a)}}{{(1 + a)(1 + c)}}$
Simplifying again we get,
$\Rightarrow \dfrac{{(1 + a)(a + ib)}}{{(1 + a)(1 + c)}}$
Eliminating the common term from both numerator and denominator we get,
$\Rightarrow \dfrac{{(a + ib)}}{{(1 + c)}}$
Hence we get,
$\dfrac{{1 + iz}}{{1 - iz}} = \dfrac{{(a + ib)}}{{(1 + c)}}$

$\therefore$ Option A is the correct answer.

Note: Imaginary numbers are defined as the square root of the negative numbers where it does not have a definite value. It is mostly written in the form of real numbers multiplied by the imaginary unit called $i$.
An imaginary number is a number that, when squared, has a negative result. Essentially, an imaginary number is the square root of a negative number and does not have a tangible value. While it is not a real number — that is, it cannot be quantified on the number line — imaginary numbers are "real" in the sense that they exist and are used in math.