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Find the number of complex numbers satisfying $|z|=z+1+2i$ .
My method: I know $|z|$ is real. So, the imaginary part of the RHS should be equal to $0$. So, $z$ should be of the form $x-2i$. Using that I am getting an imaginary value for x itself! Which, according to my method should not be. So, what is the correct method? And what is wrong with mine?

Jamil Ahmed
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    $z+1+2i$ is real, so the imaginary part of $z$ is $-2i$. Then you need to find $x$ such that $\sqrt{x^2+4}=x+1$, or $x^2+4=x^2+2x+1$... – anderstood Dec 20 '16 at 02:18
  • @anderstood my calculation probably went wrong somewhere. But, thx. – Jamil Ahmed Dec 20 '16 at 02:38
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    What is wrong is probably what you have hidden behind "Using that I am getting an imaginary value for x itself!". But you did not explain so I cannot help you... – anderstood Dec 20 '16 at 03:24
  • @anderstood I made a mistake in |z| calculation. I understood. Thank-you. – Jamil Ahmed Dec 20 '16 at 04:16

2 Answers2

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If we take $z=x+iy$ then:

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

but the RHS must be real, so $y=-2$, hence:

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

and we have to solve this equation with $x+1\ge0\longrightarrow x\ge-1$; squaring:

$$x^2+4=x^2+1+2x$$ $$x={3\over2}$$

MattG88
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Multiplying $\,z = |z|-1-2i\,$ and $\,\bar z = \overline{|z|-1-2i} = |z| - 1+2i\,$ gives:

$$|z|^2 = (|z|-1)^2+4 = |z|^2 - 2 |z| + 5$$

Therefore $\,|z|=\frac{5}{2}\,$, and substituting back into the original equation: $\,z=\frac{3}{2}-2i\,$.

dxiv
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