Question

The point of intersection the curves $arg(z-i+2)=\frac{\pi }{6}$ & $arg(z+4-3i)=-\frac{\pi }{4}$ is given by

• A. $(-2+i)$
• B. $2-i$
• C. $2+i$
• D. $N.O.T$

Answer 1

The correct option is A $(-2+i)$

$\begin{array}{c}\left(z-i+2\right)=\frac{\pi }{6}and,\left(z+4-3i\right)=\frac{-\pi }{4}\\ \left(z-\left(-2+i\right)\right)=\frac{\pi }{6}and\left(z-\left(-4+3i\right)\right)=\frac{\pi }{4}\\ Fromthefigure\\ y-3=-\left(x+4\right)\\ y+x+1=0\\ \sqrt{3}y-x-2-\sqrt{3}=0\\ -------------\\ \left(1+\sqrt{3}\right)y=\sqrt{3}+1\\ y=1\\ x=-2\\ Clearlypointofinter\sec tionis\left(-2,1\right)\\ Hence,theoptionAisthecorrectanswer.\end{array}$