Solve for all complex numbers z that satisfy the equation

Question

I am given an equation, $$|z|+z=2+i$$ I need to solve for all complex numbers z that satisfy the above equation. Should I just put $z=x+iy$, $|z|=\sqrt{x^2+y^2}$, and then solve it? But doing this I would get,

$$\sqrt{x^2+y^2}+(x+iy)=(2+i)$$ $$\implies\sqrt{x^2+y^2}=(2+i)-(x+iy)$$ Then I would have to square it, and wouldn't it then introduce extraneous roots? Also I think this seems to be pretty inefficient method for this question, is there any other way? Squaring this equation would turn pretty bad pretty fast I think.

Answer 1

From your last equation, the two members are real so that $y=1$. Then you are left with

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

If you remind that $x\le2$ you can square and get

$$x^2+1=4-4x+x^2$$ or $$4x=3.$$

Answer 2

For $z = x+iy$, you know that $y$ is the only term with an imaginary part so that $y =1$. Then we solve for $x$ by rewriting the equation as$\sqrt{x^2+1} + x =2$. Solving this gives you that $x = 3/4$.

Answer 3

Let $z=re^{i\theta}$. Then, the equation $|z|+z = 2+i$ becomes $r+re^{i\theta}=2+i$, or

$$r(1+\cos\theta) =2, \>\>\>\>\>r\sin\theta = 1$$

Note

$$ \tan\frac\theta2=\frac{\sin\theta}{1+\cos\theta} =\frac12$$

Then,

$$\sin\theta=\frac{2\tan\frac{\theta}2}{1+\tan^2\frac{\theta}2}=\frac45,\>\>\>\>\> \cos\theta=\frac{1-\tan^2\frac{\theta}2}{1+\tan^2\frac{\theta}2}=\frac35,\>\>\>\>\> r = \frac1{\sin\theta} = \frac54$$

Thus, the solution is $z=r(\cos\theta+i\sin\theta) = \frac34+i$.