<s>[INST] 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?
 [/INST] 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
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