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

Hint: 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,
\[a= \dfrac{{x - 1 + iy}}{{x + 1 + iy}} \times \dfrac{{x + 1 - iy}}{{x + 1 - iy}} \\
\Rightarrow a= \dfrac{{(x - 1 + iy)(x + 1 - iy)}}{{(x + 1 + iy)(x + 1 - iy)}} \\ \]
\[\Rightarrow a= \dfrac{{(x - (1 - iy))(x + (1 - iy))}}{{((x + 1) + iy)((x + 1) - iy)}}\]
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 \[\left| z \right| = 1\].