If - z 1- is a complex number other than -1 such that - z 1-1 and z 2-z1-1-z1-1- then show that the real parts of z 2 is zero-

Let z_1=x_1+iy_1, z_2=x_2+iy_2 |z_1|=1⇒ x_1^2+y_1^2=1 z_2=z_1 1/z_1+1 x_2+iy_2=x_1+iy_1 1/x_1+iy_1+1 ⇒ x_2+iy_2=x_1 1+iy_1/x_1+1+iy_1 ⇒ x_2+iy_2=(x_1 1+iy_1) ...

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Byju's Answer

Standard XII

Mathematics

Purely Imaginary

If | z 1| is ...

Question

If $| z_{1} |$ is a complex number other than -1 such that $| z_{1} | = 1 and z_{2} = \frac{z_{1} - 1}{z_{1} + 1}$ , then show that the real parts of $z_{2}$ is zero.

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Solution

$Let z_{1} = x_{1} + i y_{1}, z_{2} = x_{2} + i y_{2} | z_{1} | = 1 \Rightarrow x_{1}^{2} + y_{1}^{2} = 1 z_{2} = \frac{z_{1} - 1}{z_{1} + 1} x_{2} + i y_{2} = \frac{x_{1} + i y_{1} - 1}{x_{1} + i y_{1} + 1} \Rightarrow x_{2} + i y_{2} = \frac{x_{1} - 1 + i y_{1}}{x_{1} + 1 + i y_{1}} \Rightarrow x_{2} + i y_{2} = \frac{(x_{1} - 1 + i y_{1}) (x_{1} + 1 - i y_{1})}{(x_{1} + 1 + i y_{1}) (x_{1} + 1 - i y_{1})} [Rationalizing the denominator] \Rightarrow x_{2} + i y_{2} = \frac{(x_{1} - 1) (x_{1} + 1) - i y_{1} (x_{1} - 1) + i y_{1} (x_{1} + 1) + y_{1}^{2}}{(x_{1} + 1)^{2} - (i y_{1})^{2}} \Rightarrow x_{2} + i y_{2} \Rightarrow \frac{x_{1}^{2} - 1 + y_{1}^{2} - i y_{1} x_{1} + i y_{1} + i y_{1} x_{1} + i y_{1}}{(x_{1} + 1)^{2} - (i y_{1})^{2}} \Rightarrow x_{2} + i y_{2} = \frac{x_{1}^{2} + y_{1}^{2} - 1 + 2 i y_{1}}{(x_{1} + 1)^{2} - (i y_{1})^{2}} \Rightarrow x_{2} + i y_{2} = \frac{1 - 1 + 2 i y_{1}}{(x_{1} + 1)^{2} - (i y_{1})^{2}} [∵ x_{1}^{2} + y_{1}^{2} = 1] \Rightarrow x_{2} + i y_{2} = \frac{2 i y_{1}}{(x_{1} + 1)^{2} - (i y_{1})^{2}} [∵ x_{1}^{2} + y_{1}^{2} = 1]$

Since there is no real part in the RHS, therefore, $x_{2} = 0$

The real part of the $z_{2} = 0$ .

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If |z1| is a complex number other than -1 such that |z1|=1 and z2=z1−1z1+1 , then show that the real parts of z2 is zero.

If $| z_{1} |$ is a complex number other than -1 such that $| z_{1} | = 1 and z_{2} = \frac{z_{1} - 1}{z_{1} + 1}$ , then show that the real parts of $z_{2}$ is zero.