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

Name: RDSHA - Grade 11 - Mathematics - Complex Numbers - Q24
Uploaded: 2022-07-04T10:36:42+05:30
Description: ⇒ 𝑧_2 =(𝑥−1)(𝑥+1)−𝑖𝑦(𝑥−1)+𝑖𝑦(𝑥+1)−𝑖^2 𝑦^2)(𝑥+1)^2−𝑖^2 𝑦^2 ⇒ 𝑧_2=𝑥^2−1−𝑖𝑦(𝑥−1−𝑥−1)−(−1) 𝑦^2(𝑥+1)^2−(−1) 𝑦^2 _2=𝑥^2+𝑦^2−1+2𝑖𝑦𝑥^2+𝑦^2+1+2𝑥 _2=1−1+2𝑖𝑦1+1+2𝑥 nbsp; ...

⇒ 𝑧_2 =(𝑥−1)(𝑥+1)−𝑖𝑦(𝑥−1)+𝑖𝑦(𝑥+1)−𝑖^2 𝑦^2)(𝑥+1)^2−𝑖^2 𝑦^2 ⇒ 𝑧_2=𝑥^2−1−𝑖𝑦(𝑥−1−𝑥−1)−(−1) 𝑦^2(𝑥+1)^2−(−1) 𝑦^2 _2=𝑥^2+𝑦^2−1+2𝑖𝑦𝑥^2+𝑦^2+1+2𝑥 _2=1−1+2𝑖𝑦1+1+2𝑥 nbsp; ...

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Byju's Answer

Standard XII

Mathematics

Domain

If 𝑧1is a co...

Question

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

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Solution

$\Rightarrow z_{2}$
$= \frac{(x - 1) (x + 1) - i y (x - 1) + i y (x + 1) - i^{2} y^{2})}{(x + 1)^{2} - i^{2} y^{2}}$
$\Rightarrow z_{2} = \frac{x^{2} - 1 - i y (x - 1 - x - 1) - (- 1) y^{2}}{(x + 1)^{2} - (- 1) y^{2}}$
$\Rightarrow z_{2} = \frac{x^{2} + y^{2} - 1 + 2 i y}{x^{2} + y^{2} + 1 + 2 x}$
$\Rightarrow z_{2} = \frac{1 - 1 + 2 i y}{1 + 1 + 2 x} f r o m e q n (1)$
$\Rightarrow z_{2} = \frac{2 i y}{2 (1 + x)}$
$\Rightarrow z_{2} = \frac{i y}{1 + x}$
$∴ R e (z_{2}) = 0$
Hence, proved.

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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.