Question

If we have a complex number as $z(1 + a) = b + ic$ and ${a^2} + {b^2} + {c^2} = 1$ , then $\dfrac{{(1 + iz)}}{{(1 - iz)}}$ is equal to
$1)\dfrac{{(a + ib)}}{{1 + c}}$
$2)\dfrac{{(b - ic)}}{{1 + a}}$
$3)\dfrac{{(a + ic)}}{{1 + b}}$
$4)$ None of these

Answer 1

Hint: First, complex numbers are the real and imaginary combined numbers as in the form of $z = x + iy$, where x and y are the real numbers and $i$ is the imaginary.
Imaginary $i$ can be also represented into the real values only if, ${i^2} = - 1$
We will use the complex conjugation method to solve the given problem.

Complete step-by-step solution:
Since from the given that we have $z(1 + a) = b + ic$. Let us divide both equations with $1 + a$ to get the value of the z. thus we have $\dfrac{{z(1 + a)}}{{1 + a}} = \dfrac{{b + ic}}{{1 + a}}$
Canceling the common terms, we have $z = \dfrac{{b + ic}}{{1 + a}}$ .
Now we need to find the value of $\dfrac{{(1 + iz)}}{{(1 - iz)}}$ so let us substitute the value of the z which we found, then we get $\dfrac{{(1 + iz)}}{{(1 - iz)}} = \dfrac{{(1 + \dfrac{{i(b + ic)}}{{1 + a}})}}{{(1 - \dfrac{{i(b + ic)}}{{1 + a}})}}$
and we will cross multiply the denominator values then we get $\dfrac{{(1 + iz)}}{{(1 - iz)}} = \dfrac{{(1 + \dfrac{{i(b + ic)}}{{1 + a}})}}{{(1 - \dfrac{{i(b + ic)}}{{1 + a}})}} = \dfrac{{\dfrac{{1 + a + i(b + ic)}}{{1 + a}}}}{{\dfrac{{1 + a - i(b + ic)}}{{1 + a}}}}$
now we will cancel the common terms, then we have $\dfrac{{\dfrac{{1 + a + i(b + ic)}}{{1 + a}}}}{{\dfrac{{1 + a - i(b + ic)}}{{1 + a}}}} = \dfrac{{1 + a + i(b + ic)}}{{1 + a - i(b + ic)}}$
further solving we have $\dfrac{{1 + a + i(b + ic)}}{{1 + a - i(b + ic)}} = \dfrac{{1 + a + ib + {i^2}c}}{{1 + a - ib - {i^2}c}}$ since we know that ${i^2} = - 1$
hence, we get $\dfrac{{1 + a + ib + {i^2}c}}{{1 + a - ib - {i^2}c}} = \dfrac{{1 + a - c + ib}}{{1 + a + c - ib}}$
Now we will use the method of conjugation, which is to multiply with the original denominator of the opposite sign imaginary value to the numerator and denominator values.
Thus, we have $\dfrac{{1 + a - c + ib}}{{1 + a + c - ib}} \times \dfrac{{1 + a + c + ib}}{{1 + a + c + ib}}$
Since $(a - ib)(a + ib) = {a^2} + {b^2}$ then we have $\dfrac{{1 + a - c + ib}}{{1 + a + c - ib}} \times \dfrac{{1 + a + c + ib}}{{1 + a + c + ib}} = \dfrac{{(1 + a - c + ib)(1 + a + c + ib)}}{{{{(1 + a + c)}^2} + {b^2}}}$
Further solving with the general multiplication, we have $\dfrac{{(1 + a - c + ib)(1 + a + c + ib)}}{{{{(1 + a + c)}^2} + {b^2}}} = \dfrac{{1 + 2a + {a^2} + {b^2} + {c^2} + 2ib + 2iab}}{{1 + {a^2} + {b^2} + {c^2} + 2ac + 2(a + c)}}$
Since from the given we have ${a^2} + {b^2} + {c^2} = 1$ and thus we get $\dfrac{{1 + 2a + {a^2} + {b^2} + {c^2} + 2ib + 2iab}}{{1 + {a^2} + {b^2} + {c^2} + 2ac + 2(a + c)}} = \dfrac{{2(a + {a^2} + ib + iab)}}{{2(1 + ac + (a + c))}}$
Canceling the common terms, we have $\dfrac{{(a + {a^2} + ib + iab)}}{{(1 + ac + (a + c))}} = \dfrac{{a(a + 1) + ib(a + 1)}}{{(a + 1)(c + 1)}}$
Finally, we have \[\dfrac{{a(a + 1) + ib(a + 1)}}{{(a + 1)(c + 1)}} = \dfrac{{a + ib}}{{c + 1}}\]
Therefore, the option $1)\dfrac{{(a + ib)}}{{1 + c}}$ is correct.

Note: The conjugate of a complex number represents the reflection of that complex number about the real axis on the argand plane.
When the imaginary $i$ of the complex number is replaced with $ - i$ , we get the conjugate of that complex number that shows the image of the particular complex number about the plane.
Hence, we used the conjugation of $1 + a + c - ib$ is $1 + a + c + ib$