The amplitude of left( frac{1 - i}{1 + i} rig | vedclass

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Question

(A) Let $z = \frac{1 - i}{1 + i}$.To simplify,multiply the numerator and denominator by the conjugate of the denominator $(1 - i)$:$z = \frac{1 - i}{1

Vedclass · Accepted Answer

$-\frac{\pi}{2}$

Vedclass · Answer

(A) Let $z = \frac{1 - i}{1 + i}$.To simplify,multiply the numerator and denominator by the conjugate of the denominator $(1 - i)$:$z = \frac{1 - i}{1 + i} 	imes \frac{1 - i}{1 - i} = \frac{(1 - i)^2}{1^2 - i^2} = \frac{1 - 2i + i^2}{1 - (-1)} = \frac{1 - 2i - 1}{2} = \frac{-2i}{2} = -i$.The complex number is $z = 0 - i$.Since the real part is $0$ and the imaginary part is $-1$,the point lies on the negative imaginary axis.The amplitude of a complex number $z = x + iy$ where $x=0$ and $y < 0$ is $-\frac{\pi}{2}$.

The amplitude of $\left( \frac{1 - i}{1 + i} \right)$ is

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