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z^(')=(1+ix)/(1-ix)

z′=1+ix1−ix z^{\prime}=\frac{1+i x}{1-i x}

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Full solution

Q. z′=1+ix1−ix z^{\prime}=\frac{1+i x}{1-i x}
  1. Rationalize Denominator: To solve for z′z', we need to rationalize the denominator by multiplying the numerator and the denominator by the conjugate of the denominator, which is (1+ix)(1+ix).
  2. Multiplication of Conjugates: So we multiply (1+ix)(1+ix) by (1+ix)(1+ix) for the numerator and (1−ix)(1-ix) by (1+ix)(1+ix) for the denominator.
  3. Numerator Calculation: The numerator becomes (1+ix)(1+ix)=1+2ix−x2(1+ix)(1+ix) = 1 + 2ix - x^2.
  4. Denominator Calculation: The denominator becomes (1−ix)(1+ix)=1−x2(1-ix)(1+ix) = 1 - x^2.
  5. Simplify Numerator and Denominator: Now we simplify the numerator and denominator. The numerator is 1+2ix−x21 + 2ix - x^2 and the denominator is 1−x21 - x^2.
  6. Final Expression: We can now write z′z' as 1+2ix−x21−x2\frac{1 + 2ix - x^2}{1 - x^2}.
  7. Correct Denominator Calculation: But wait, there's a mistake in the calculation of the denominator. It should be (1−ix)(1+ix)=1−i2x2(1-i x)(1+i x) = 1 - i^2 x^2, and since i2=−1i^2 = -1, the denominator is actually 1+x21 + x^2.

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