Test Index

VITEEE 2010 Paper

Question : 86 of 120

Marks: +1, -0

If z satisfies the equation |z| - z =1+2i, then z is equal to

Solution:

Given : |z| - z =1+2i

If z = x + iy, then this equation reduces to

| x + iy | - (x + iy) = 1 + 2i

⇒

(\sqrt{x^{2}+y^{2}}-x)+(-iy)=1+2i

On comparing real and imaginary parts of both sides of this equation, we get

\sqrt{x^{2}+y^{2}}-x=1

⇒

\sqrt{x^{2}+y^{2}}=1+x

\Rightarrow x^{2}+y^{2}=(1+x)^{2}

\Rightarrow x^{2}+y^{2}=1+x^{2}+2x

\Rightarrow y^{2}=1+2x

...(i)

and -y = 2 ⇒ y = -2

Putting this value in eq.(i), we get

(-2)^{2}=1+2x

⇒

2x=3\Rightarrow x=\frac{3}{2}

\therefore\; z=x+iy=\frac{3}{2}-2i

Go to Question:

Prev Question

Next Question