Question

The point of intersection of the curves $arg(z-3i)=\frac{3\pi }{4}$ and $arg(2z+1-2i)=\frac{\pi }{4}$, (where $i=\sqrt{-1}$) is

• A. $\frac{1}{4}(3+9i)$
• B. $\frac{1}{4}(3-9i)$
• C. $\frac{1}{2}(3+2i)$
• D. $Nopoint$

Answer 1

Answer: (A)

$\frac{1}{4}(3+9i)$

According to question,

$\begin{array}{c}(z-3i)=\frac{3\pi }{4}\\ \Rightarrow (x+i(y-3\left)\right)=\frac{3\pi }{4}\\ \Rightarrow {\tan }^{-1}\left(\frac{y-3}{x}\right)=\frac{3\pi }{4}\\ \Rightarrow \tan \frac{3\pi }{4}=\frac{y-3}{x}\\ \Rightarrow \frac{y-3}{x}=-1\\ \Rightarrow \left(x+y\right)=1-----\left(1\right)\\ and,\\ arg(2z+1-2i)=\frac{\pi }{4}\\ \Rightarrow \left(2\right(x+iy)+1-2i)=\frac{\pi }{4}\\ \Rightarrow (2x+1+i(2y-2\left)\right)=\frac{\pi }{4}\\ \Rightarrow {\tan }^{-1}\left(\frac{2y-2}{2x+1}\right)=\frac{\pi }{4}\\ \Rightarrow \frac{2y-2}{2x+1}=1\\ \Rightarrow 2y-2=2x+1\\ \Rightarrow 2x-2y=-3\\ \therefore x-y=-\frac{3}{2}--------\left(2\right)\\ solve,x=\frac{3}{4},y=\frac{9}{4}\\ \therefore P.Ix=iy=\frac{1}{4}(3+9i)\\ So,thatthecorrectoptionisA.\end{array}$