Question

The modulus of the complex number $ z $ such that $ |z+3-i|=1 $ and $ arg\,(z)=\pi $ is equal to

• A. 1
• B. 2
• C. 9
• D. 4
• E. 3

Answer 1

Answer: (E)

Let $ z=x+iy $
$ \therefore $ $ |z+3-i|=1 $
$ \Rightarrow $ $ |x+iy+3-i|=1 $
$ \Rightarrow $ $ |x+3+i(y-1)|=1 $
$ \Rightarrow $ $ {{(x+3)}^{2}}+{{(y-1)}^{2}}=1 $ ...(i) But $ {{\tan }^{-1}}\frac{y}{x}=\pi $
$ \Rightarrow $ $ y=0 $ ...(ii)
$ \therefore $ From Eqs. (i) and (ii), we get $ {{(x+3)}^{2}}=0 $
$ \Rightarrow $ $ x=-3,-3 $
$ \therefore $ Required complex number is $ x+iy=-3 $
$ \therefore $ $ |x+iy|=|-3|=3 $