Question

If $a=\dfrac{{z - 1}}{{z + 1}}$ is purely imaginary number (z is not equal to$- 1$), then mod $z$ is ?
A. $1$
B. $2$
C. $3$
D. $4$

Answer 1

A complex number is of the form a +ib where a and b are real numbers and $i$ is an imaginary number. Here, we will take $z = x + iy$ and substitute the value of z. We will also use the formula $(a - b)(a + b) = {a^2} - {b^2}$ and ${(a - b)^2} = {a^2} + {b^2} - 2ab$. We know that ${i^2} = - 1$. For the given complex number is in the form $z = x + iy$, then the conjugate of that number will be $\overline z = x - iy$. Here, if you take the conjugate number as a complex number and then solve it, you will still get the correct answer.

Complete step by step answer:
Let, $z = x + iy$ and $z \ne - 1$.
Give that, $a = \dfrac{{z - 1}}{{z + 1}}$
Substituting the value of z, we get,
$a = \dfrac{{x + iy - 1}}{{x + iy + 1}}$
Rearrange the above expression, we get,
$a = \dfrac{{x - 1 + iy}}{{x + 1 + iy}}$
Multiply to both the numerator and denominator with$x + 1 - iy$, we get,

\Rightarrow a= \dfrac{{(x - 1 + iy)(x + 1 - iy)}}{{(x + 1 + iy)(x + 1 - iy)}} \\\ $$ $$\Rightarrow a= \dfrac{{(x - (1 - iy))(x + (1 - iy))}}{{((x + 1) + iy)((x + 1) - iy)}}$$ Applying this formula$$(a - b)(a + b) = {a^2} - {b^2}$$, we get, $$a = \dfrac{{{x^2} - {{(1 - iy)}^2}}}{{{{(x + 1)}^2} - {{(iy)}^2}}}$$ Applying this formula$${(a - b)^2} = {a^2} + {b^2} - 2ab$$, we get, Also, $${(iy)^2} = {i^2}{y^2} = ( - 1){y^2} = - {y^2}$$ $$(\because {i^2} = - 1)$$ $$ a= \dfrac{{{x^2} - ({1^2} + {{(iy)}^2} - 2(iy))}}{{{{(x + 1)}^2} - {{(iy)}^2}}}$$ $$\Rightarrow a = \dfrac{{{x^2} - (1 - {y^2} - 2yi)}}{{{{(x + 1)}^2} - ( - {y^2})}}$$ $$\Rightarrow a = \dfrac{{{x^2} - 1 + {y^2} + 2yi}}{{{{(x + 1)}^2} + {y^2}}}$$ $$\Rightarrow a = \dfrac{{{x^2} + {y^2} - 1}}{{{{(x + 1)}^2} + {y^2}}} + \dfrac{{2y}}{{{{(x + 1)}^2} + {y^2}}}i$$ Since it is given that $$\dfrac{{z - 1}}{{z + 1}}$$ is purely imaginary, it means that the real part is zero. $$\dfrac{{{x^2} + {y^2} - 1}}{{{{(x + 1)}^2} + {y^2}}} = 0$$ $$\Rightarrow {x^2} + {y^2} - 1 = 0 \\\ \Rightarrow {x^2} + {y^2} = 1 \\\ $$ We know that, $$\left| z \right| = \left| {x + iy} \right| = \sqrt {{x^2} + {y^2}} $$ $$ \Rightarrow \sqrt {{x^2} + {y^2}} = \sqrt 1 $$ $$ \Rightarrow \left| z \right| = \pm 1$$ Since, it is given that, $$z \ne - 1$$ $$ \therefore \left| z \right| = 1$$ **Note:** Another method: Since it is given that$$\dfrac{{z - 1}}{{z + 1}}$$is purely imaginary, $$\dfrac{{z - 1}}{{z + 1}} + (\overline {\dfrac{{z - 1}}{{z + 1}}} ) = 0$$ $$ \Rightarrow \dfrac{{z - 1}}{{z + 1}} + \dfrac{{\overline z - 1}}{{\overline z + 1}} = 0$$ $$ \Rightarrow \dfrac{{(z - 1)(\overline z + 1) + \overline {(z} - 1)(z + 1)}}{{(z + 1)(\overline z + 1)}} = 0$$ $$ \Rightarrow (z - 1)(\overline z + 1) + \overline {(z} - 1)(z + 1) = 0$$ $$\Rightarrow z\overline z + z - \overline z - 1 + \overline z z + \overline z - z - 1 = 0 \\\ \Rightarrow 2z\overline z - 2 = 0 \\\ $$ $$ \Rightarrow z\overline z - 1 = 0$$ $$ \Rightarrow {\left| z \right|^2} = 1$$ $$ \Rightarrow \left| z \right| = \pm 1$$ Since, it is given that, $$z \ne - 1$$ $$ \therefore \left| z \right| = 1$$ Hence, if $$\dfrac{{z - 1}}{{z + 1}}$$ is purely imaginary number and$$z \ne - 1$$ then $$\left| z \right| = 1$$.