The polar form of the complex number z = frac | vedclass

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Question

(A) Step $1$: Simplify $z = \frac{1}{1+i}$ by multiplying the numerator and denominator by the conjugate $(1-i)$.$z = \frac{1(1-i)}{(1+i)(1-i)} = \fra

Vedclass · Accepted Answer

$\frac{1}{\sqrt{2}} (\cos \frac{7\pi}{4} + i\sin \frac{7\pi}{4})$

Vedclass · Answer

(A) Step $1$: Simplify $z = \frac{1}{1+i}$ by multiplying the numerator and denominator by the conjugate $(1-i)$.$z = \frac{1(1-i)}{(1+i)(1-i)} = \frac{1-i}{1^2 - i^2} = \frac{1-i}{1+1} = \frac{1}{2} - \frac{1}{2}i$.Step $2$: Find the modulus $r = |z| = \sqrt{(\frac{1}{2})^2 + (-\frac{1}{2})^2} = \sqrt{\frac{1}{4} + \frac{1}{4}} = \sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}}$.Step $3$: Find the argument $	heta$. Since $z$ is in the fourth quadrant $(x > 0, y Thus, $	heta = 2\pi - \frac{\pi}{4} = \frac{7\pi}{4}$.Step $4$: The polar form is $z = r(\cos 	heta + i\sin 	heta) = \frac{1}{\sqrt{2}}(\cos \frac{7\pi}{4} + i\sin \frac{7\pi}{4})$.

The polar form of the complex number $z = \frac{1}{1 + i}$, (where $i = \sqrt{-1}$) is

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