Find couples of complex numbers

Question

I found this exercise, given: $$u=|z|+|u|$$ and $$z=|u|+1$$ (it is a system I don't how to write it in latex from) I have to find the couples of complex numbers $u,z$ that comes from the two equation. My first attempt was following algebraic way, the second attempt was the trigonometric way but in both case I came to equation that I don't know how to resolve. Probably there is a simpler way, can you help me please? Thank you in advance.

Answer 1

From the first equation, since absolute values are real and non-negative, you can derive that $u\in\mathbb R$ and $u\geq0$. This implies that $u=\lvert u\rvert$, hence the second equation becomes $z=u+1$, yielding $z\in\mathbb R$ and $z\geq0$. Substitute $z$ in the first equation to get $$u=(u+1)+u\text,$$ whose only solution is $u=-1$. But this contradicts $u\geq0$, so the system cannot be solved.

Answer 2

If you wrote that accurately then the numbers are real non-complex:

$$u=a+bi\;,\;\;z=x+yi\implies\begin{cases}a+bi=|z|+|u|\implies a= |z|+|u|\;,\;\;b=0\\{}\\x+yi=1+|u|\implies x=1+|u|\;,\;\;y=0\end{cases}$$

and then

$$a=|z|+|u|=(1+|u|)+|u|=1+2|u|=1+2|a|\implies a-2|a|=1\implies\begin{cases}a-2a=1\iff a=-1\;,\;\;\text{if}\;\;\;\;a\ge 0\\{}\\a+2a=1\iff a=\frac13\;,\;\;\text{if}\;\;\; a<0\end{cases}$$

and we get contradictions.

Answer 3

the apparent triviality of the problem as posed suggests there may have been a copying error? the following version seems to offer a more appropriate elementary question concerning the geometry of complex numbers: $$ |z+u|=|u| \\ |z|=|u+1| $$