The sum of the argument of z and another comp | vedclass

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(B) Let $z = r(\cos 	heta + i \sin 	heta)$,where $	heta = \arg(z)$.Let the other complex number be $w = R(\cos \phi + i \sin \phi)$,where $\phi = \

Vedclass · Accepted Answer

$-\bar{z}$

Vedclass · Answer

(B) Let $z = r(\cos 	heta + i \sin 	heta)$,where $	heta = \arg(z)$.Let the other complex number be $w = R(\cos \phi + i \sin \phi)$,where $\phi = \arg(w)$.Given that $\arg(z) + \arg(w) = \pi$,we have $	heta + \phi = \pi$,which implies $\phi = \pi - 	heta$.Thus,$w = R(\cos(\pi - 	heta) + i \sin(\pi - 	heta)) = R(-\cos 	heta + i \sin 	heta)$.If we consider $z = x + iy$,then $\bar{z} = x - iy = r(\cos 	heta - i \sin 	heta)$.Then $-\bar{z} = -x + iy = r(-\cos 	heta + i \sin 	heta)$.Comparing this with $w$,we see that $w = -\bar{z}$ (assuming $R=r$).

The sum of the argument of $z$ and another complex number is $\pi$. The other complex number can be written as:

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