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Question: If \({z_1}\) is a complex number other than \( - 1\) such that \(\left| {{z_1}} \right| = 1\) and \(...

If z1{z_1} is a complex number other than −1 - 1 such that ∣z1∣=1\left| {{z_1}} \right| = 1 and z2=z1−1z1+1{z_2} = \dfrac{{{z_1} - 1}}{{{z_1} + 1}}, then show that the real parts of z2{z_2} is zero.

Explanation

Solution

As in the question we have given z1{z_1} is a complex number let’s consider z1=x+iy{z_1} = x + iy where xx is real part and iyiy is imaginary part. Using all the given conditions which are given in the question we can find the z2{z_2} and arrive at the required answer.

Complete step by step answer:
Here in this question we have given that z1{z_1} is a complex number, so we can write z1{z_1} as z1=x+iy{z_1} = x + iy where xx is real part and iyiy is imaginary part.
As we have given ∣z1∣=1\left| {{z_1}} \right| = 1 we can write as below.
∣z1∣=1\left| {{z_1}} \right| = 1
⇒x2+y2=1\Rightarrow \sqrt {{x^2} + {y^2}} = 1
Squaring on both the sides and simplifying, we get
⇒x2+y2=1\Rightarrow {x^2} + {y^2} = 1
Also we have given z1≠−1{z_1} \ne - 1
⇒x+iy≠−1\Rightarrow x + iy \ne - 1
⇒x≠−1\Rightarrow x \ne - 1
Now, they have asked to show the real part of z2{z_2} is zero in the expression z2=z1−1z1+1{z_2} = \dfrac{{{z_1} - 1}}{{{z_1} + 1}}.
Now, in the above expression of z2{z_2} replace z1=x+iy{z_1} = x + iy and simplify the expression. Therefore, we get
z2=(x+iy)−1(x+iy)+1{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.
⇒z2=(x−1)+iy(x+1)+iy\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.
⇒z2=(x−1)+iy(x+1)+iy×(x−1)−iy(x+1)−iy\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
⇒z2=(x−1)(x+1)−(x−1)y.i+(x+1)y.i−y2i2(x2+1)2−(iy)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}}}
⇒z2=(x−1)(x+1)−(x−1)y.i+(x+1)y.i−y2i2(x2+1)2−(iy)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
⇒z2=(x2−1+y2)+i(−xy+y+xy+y)(x+1)2+y2\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 i2=−1{i^2} = - 1)
⇒z2=(x2+y2−1)+2iy(x+1)2+y2\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 x2+y2=1{x^2} + {y^2} = 1 in the above equation, we get
⇒z2=(1−1)+2iy(x+1)2+y2\Rightarrow {z_2} = \dfrac{{\left( {1 - 1} \right) + 2iy}}{{{{\left( {x + 1} \right)}^2} + {y^2}}}
⇒z2=2iy(x+1)2+y2\Rightarrow {z_2} = \dfrac{{2iy}}{{{{\left( {x + 1} \right)}^2} + {y^2}}}
The above expression can be written as
⇒z2=0+i.2y(x+1)2+y2\Rightarrow {z_2} = 0 + i.\dfrac{{2y}}{{{{\left( {x + 1} \right)}^2} + {y^2}}}
In the above expression we have the real part as 00 and the imaginary part as i.2y(x+1)2+y2i.\dfrac{{2y}}{{{{\left( {x + 1} \right)}^2} + {y^2}}}.

Hence we have proved that the real parts of z2{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.