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Q.
Point of intersection of the lines arg $(z-1)=\frac{\pi}{4}$ and $z=i t+(1-t)$ is $z=i t+(1-t)$ is
Complex Numbers and Quadratic Equations
Solution:
$\arg (z-1)=\frac{\pi}{4}$
i.e., a ray with initial point $(1,0)$ however excluding this point and inclined at an angle $45^{\circ}$ with $x$-axis.
$z=i t+(1-t)$
i.e., $x=1-t$ and $y=t$
or $x+y=1$
Clearly, point of intersection is $(1,0)$.
But $(1,0)$ does not satisfy the first equation.
Hence, no point of intersection.