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If ${z_1}$ is a complex number other than $ - 1$ such that $\left| {{z_1}} \right| = 1$ and ${z_2} = \dfrac{{{z_1} - 1}}{{{z_1} + 1}}$, then show that the real parts of ${z_2}$ is zero.

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Last updated date: 04th Mar 2024
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IVSAT 2024
Answer
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Hint: As in the question we have given ${z_1}$ is a complex number let’s consider ${z_1} = x + iy$ where $x$ is real part and $iy$ is imaginary part. Using all the given conditions which are given in the question we can find the ${z_2}$ and arrive at the required answer.

Complete step by step answer:
Here in this question we have given that ${z_1}$ is a complex number, so we can write ${z_1}$ as ${z_1} = x + iy$ where $x$ is real part and $iy$ is imaginary part.
As we have given $\left| {{z_1}} \right| = 1$ we can write as below.
$\left| {{z_1}} \right| = 1$
$ \Rightarrow \sqrt {{x^2} + {y^2}} = 1$
Squaring on both the sides and simplifying, we get
$ \Rightarrow {x^2} + {y^2} = 1$
Also we have given ${z_1} \ne - 1$
$ \Rightarrow x + iy \ne - 1$
$ \Rightarrow x \ne - 1$
Now, they have asked to show the real part of ${z_2}$ is zero in the expression ${z_2} = \dfrac{{{z_1} - 1}}{{{z_1} + 1}}$.
Now, in the above expression of ${z_2}$ replace ${z_1} = x + iy$ and simplify the expression. Therefore, we get
${z_2} = \dfrac{{\left( {x + iy} \right) - 1}}{{\left( {x + iy} \right) + 1}}$
The above expression can be written as below, for the simplification purpose.
$ \Rightarrow {z_2} = \dfrac{{\left( {x - 1} \right) + iy}}{{\left( {x + 1} \right) + iy}}$
Now multiply and divide the above expression by conjugate of the value, which is as below.
$ \Rightarrow {z_2} = \dfrac{{\left( {x - 1} \right) + iy}}{{\left( {x + 1} \right) + iy}} \times \dfrac{{\left( {x - 1} \right) - iy}}{{\left( {x + 1} \right) - iy}}$
Now, we simplify the above expression. We get
$ \Rightarrow {z_2} = \dfrac{{\left( {x - 1} \right)\left( {x + 1} \right) - \left( {x - 1} \right)y.i + \left( {x + 1} \right)y.i - {y^2}{i^2}}}{{{{\left( {{x^2} + 1} \right)}^2} - {{\left( {iy} \right)}^2}}}$
$ \Rightarrow {z_2} = \dfrac{{\left( {x - 1} \right)\left( {x + 1} \right) - \left( {x - 1} \right)y.i + \left( {x + 1} \right)y.i - {y^2}{i^2}}}{{{{\left( {{x^2} + 1} \right)}^2} - {{\left( {iy} \right)}^2}}}$
Now, simplify the above expression by separating the real and imaginary terms, we write as
$ \Rightarrow {z_2} = \dfrac{{\left( {{x^2} - 1 + {y^2}} \right) + i\left( { - xy + y + xy + y} \right)}}{{{{\left( {x + 1} \right)}^2} + {y^2}}}$ (we know ${i^2} = - 1$)
$ \Rightarrow {z_2} = \dfrac{{\left( {{x^2} + {y^2} - 1} \right) + 2iy}}{{{{\left( {x + 1} \right)}^2} + {y^2}}}$
By using the condition which we have ${x^2} + {y^2} = 1$ in the above equation, we get
$ \Rightarrow {z_2} = \dfrac{{\left( {1 - 1} \right) + 2iy}}{{{{\left( {x + 1} \right)}^2} + {y^2}}}$
$ \Rightarrow {z_2} = \dfrac{{2iy}}{{{{\left( {x + 1} \right)}^2} + {y^2}}}$
The above expression can be written as
$ \Rightarrow {z_2} = 0 + i.\dfrac{{2y}}{{{{\left( {x + 1} \right)}^2} + {y^2}}}$
In the above expression we have the real part as $0$ and the imaginary part as $i.\dfrac{{2y}}{{{{\left( {x + 1} \right)}^2} + {y^2}}}$.

Hence we have proved that the real parts of ${z_2}$ are zero.

Note:
Whenever we have this type of problem, first we need to consider complex values and then when it comes to the simplification part, it’s very important to be careful while solving using the conjugate values of the corresponding values. If you fail to take the correct conjugate values then you may end up with the wrong answer.
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