If arg(z - a) = frac{pi}{4},where a in R,then | vedclass

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(D) Let $z = x + iy$. Then $z - a = (x - a) + iy$.Given $arg(z - a) = \frac{\pi}{4}$.This implies $	an^{-1}\left(\frac{y}{x - a}ight) = \frac{\pi}{

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Straight line

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(D) Let $z = x + iy$. Then $z - a = (x - a) + iy$.Given $arg(z - a) = \frac{\pi}{4}$.This implies $	an^{-1}\left(\frac{y}{x - a}ight) = \frac{\pi}{4}$ for $x > a$.Therefore,$\frac{y}{x - a} = 	an\left(\frac{\pi}{4}ight) = 1$.This simplifies to $y = x - a$,or $x - y - a = 0$,which represents a ray starting from $(a, 0)$ excluding the point $(a, 0)$ itself.Since this is a part of a line,the locus is a straight line.

If $arg(z - a) = \frac{\pi}{4}$,where $a \in R$,then the locus of $z \in C$ is a

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