Does $\rvert z \rvert +z= i$ have any solutions? $$z=a+i \ b$$
$$\vert z \rvert = \sqrt{a^2+b^2} \\$$
$$\sqrt{a^2+b^2}+a+i \ b=i \\$$
\begin{cases}b=1 \\
\sqrt{a^2+b^2}+a=0 \end{cases}

$$\sqrt{a^2+1}+a=0$$
$$\sqrt{a^2+1}=-a$$
$$a^2+1=a^2 \\ \\$$
The equation has no solution
Is it correct?
 A: There are some good answers here, but I'll go ahead and add mine.
If $|z|+z=i$, then because our real parts must cancel and $|z|$ is real, we conclude $\rm{Re}(z)=-|z|$; however, because 
$$|z|^2=(\rm{Re}\:z)^2+(\rm{Im}\:z)^2=|z|^2+(\rm{Im}\:z)^2$$
we conclude $\text{Im}(z)=0$. But then $|z|+z=i$ would have zero imaginary part, which is a contradiction.
A: with $$z=a+bi$$ you will get the equation
$$\sqrt{a^2+b^2}+a+bi-i=0$$ this is equivalent to
$$\sqrt{a^2+b^2}+a+i(b-1)=0$$ from here we get
$$b=1$$
and $$\sqrt{a^2+b^2}+a=0$$
from the second equation we get
$$\sqrt{a^2+b^2}=-a$$
squaring gives 
$$a^2+b^2=a^2$$ substracting $a^2$ gives $$b=0$$
this is a contradiction to $$b=1$$
thus the given equation does not hold
A: There's a much easier way to see this. Think of complex numbers as points in the plane. The number $i$ is the point on the $y$ axis one unit from the origin.
We want to find a point $z$ which, translated to the right by a certain amount, moves exactly to $i$. This is because we are adding a positive real number to $z$, and adding a positive real number always translates to the right.
So the point must lie on the horizontal line through $i$. This is your $b=1$ equation.
Now how far does $z$ move to the right? $|z|$ is the distance from $z$ to 0. But this distance is always a little more than the distance from $z$ to $i$ because from $z$ to $i$ is straight along the line but from $z$ to $0$ is the same distance to the right plus one unit down. 
So no matter where $z$ is, if we move $z$ rightward by $|z|$ we always move it too far to get it exactly to $i$.

