Question

find the no. of complex numbers satisfying |z|=z+1+2i

Answer 1

Answer:

Step-by-step explanation:

Let, z=x+iy where x and y are real numbers.So,|z|=z+1+2i⇒x2+y2−−−−−−√=x+iy+1+2i⇒x2+y2−−−−−−√=(x+1)+i(y+2)On comparing the real and imaginary parts, we have,y+2=0⇒y=−2And,x2+y2−−−−−−√=x+1⇒x2+4−−−−−√=x+1 (put y=−2)Squaring on both side,x2+4=x2+1+2x⇒2x=3⇒x=32Thusz=3/2−2i is the solution of given equation.

Answer 2

The no. of complex numbers satisfying |z|=z+1+2i is z=3/2-2i.

Step-by-step explanation:

Given:

Let, z=x+iy where x and y are real numbers.

So,

|z|=z+1+2i

$\sqrt{x^2+y^2}=x+iy+1+2i$

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

On comparing the real and imaginary parts, we have,

y+2=0

⇒y=−2

And,

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

$\sqrt{x^2+4}=x+1$                   (put y=−2)

Squaring on both side, we get

$x^2+4=x^2+1+2x$

2x=4-1

2x=3

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

Thus $z=\frac{3}{2}-2i$ is the solution of given equation.