Find the modulus and argument of the complex | vedclass

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(A) Let $z = \frac{1}{1+i}$.Multiplying the numerator and denominator by the conjugate $(1-i)$,we get:$z = \frac{1(1-i)}{(1+i)(1-i)} = \frac{1-i}{1^2

Vedclass · Accepted Answer

Modulus = $\frac{1}{\sqrt{2}}$,Argument = $\frac{-\pi}{4}$

Vedclass · Answer

(A) Let $z = \frac{1}{1+i}$.Multiplying the numerator and denominator by the conjugate $(1-i)$,we get:$z = \frac{1(1-i)}{(1+i)(1-i)} = \frac{1-i}{1^2 - i^2} = \frac{1-i}{1+1} = \frac{1}{2} - \frac{1}{2}i$.Here,the real part $x = \frac{1}{2}$ and the imaginary part $y = -\frac{1}{2}$.The modulus $r = |z| = \sqrt{x^2 + y^2} = \sqrt{(\frac{1}{2})^2 + (-\frac{1}{2})^2} = \sqrt{\frac{1}{4} + \frac{1}{4}} = \sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}}$.For the argument $	heta$,since $x > 0$ and $y $	an 	heta = |\frac{y}{x}| = |\frac{-1/2}{1/2}| = |-1| = 1$.Since $	heta$ is in the fourth quadrant,$	heta = -	an^{-1}(1) = -\frac{\pi}{4}$.Thus,the modulus is $\frac{1}{\sqrt{2}}$ and the argument is $-\frac{\pi}{4}$.

Find the modulus and argument of the complex number: $\frac{1}{1+i}$

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