The argument and modulus of frac{1 + i}{1 - i | vedclass

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Question

(D) Let $z = \frac{1 + i}{1 - i}$.Multiplying the numerator and denominator by the conjugate of the denominator $(1 + i)$:$z = \frac{(1 + i)(1 + i)}{(

Vedclass · Accepted Answer

$\frac{\pi}{2}$ and $1$

Vedclass · Answer

(D) Let $z = \frac{1 + i}{1 - i}$.Multiplying the numerator and denominator by the conjugate of the denominator $(1 + i)$:$z = \frac{(1 + i)(1 + i)}{(1 - i)(1 + i)} = \frac{1 + 2i + i^2}{1^2 - i^2} = \frac{1 + 2i - 1}{1 + 1} = \frac{2i}{2} = i$.We can write $z = 0 + 1i$.The modulus is $|z| = \sqrt{0^2 + 1^2} = 1$.The argument $	heta$ satisfies $\cos 	heta = 0$ and $\sin 	heta = 1$,which gives $	heta = \frac{\pi}{2}$.Thus,the argument is $\frac{\pi}{2}$ and the modulus is $1$.

The argument and modulus of $\frac{1 + i}{1 - i}$ are respectively:

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