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Question
If |z1| = 1(z1 ≠ –1) and z2 = `(z_1 - 1)/(z_1 + 1)`, then show that the real part of z2 is zero.
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Solution
Let z1 = x + yi
|z1| = `sqrt(x^2 + y^2)` = 1 ......[Given that |z1| = 1]
⇒ x2 + y2 = 1 ......(i)
Now z2 = `(z_1 - 1)/(z_1 + 1)`
= `(x + yi - 1)/(x + yi + 1)`
= `((x + 1) + y"i")/((x + 1) + y"i")`
= `((x - 1) + yi)/((x + 1) + yi) xx (x + 1 - yi)/(x + 1 - yi)`
= `((x - 1)(x + 1) - y(x - 1)i + y(x + 1)i - y^2i^2)/((x + 1)^2 - y^2i^2)`
= `(x^2 - 1 + yi(x + 1 - x + 1) + y^2)/(x^2 + 1 + 2x + y^2)`
= `((x^2 + y^2 - 1) + 2yi)/(x^2 + y^2 + 2x + 1)`
= `((1 - 1))/(x^2 + y^2 + 2x + 1) + (2y)/(x^2 + y^2 + 2x + 1) "i"`
= `0 + (2y)/(x^2 + y^2 + 2x + 1) "i"`
Hence, the real part of z2 is 0.
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