I have been practicing some complex numbers and came across this problem.
Find all the complex numbers z that satisfie the equation $z + |z| = 8+4i$
I have said that $z = x + yi$ which means $|z| = \sqrt{x^2+y^2}$ which makes the equation
$x + yi + \sqrt{x^2 +y^2} = 8 + 4i$ but I dont know how to go from here to a solution?
Do I just have to rearrange a little or am I completely barking up the wrong tree?
On
You have that $$\sqrt{x^2+y^2}+x-8=(4-y)i.$$ The LHS is real. So, the RHS must be real. Thus $y=4.$ Now you only need to solve
$$\sqrt{x^2+4^2}+x-8=0.$$
On
From $z + |z| = 8+4i$ and $z=x+iy$ we get $|z|=8-x+i(4-y) \in \mathbb R$, hence $y=4$ , $|z|=8-x$ and $z=x+4i$. Thus
$$(8-x)^2=|z|^2=x^2+16.$$
This gives $x=3$ and $z=3+4i$.
On
Notice that you would have been able to go further if you had been precise in what you tried:
Let $x,y$ be reals such that $z = x+yi$.
You can do this because this is how complex numbers are defined. And then it should immediately become obvious what to do next after you got your expression:
$x+yi+\sqrt{x^2+y^2} = 8+4i$.
Again, by definition of complex numbers, two complex numbers are equal only if both real and imaginary parts match. That is why we rearrange and isolate real/imaginary parts (based on the fact that $x,y$ are real):
$(x+\sqrt{x^2+y^2}-8)+(y-4)i = 0+0i$.
And then the rest is easy; you get two equations, one for the real part and one for the imaginary part.
You can alternatively compare directly as done in other answers, but you should know and understand that when doing so you are effectively doing the same, or equivalently using the following properties:
$Re(z+w) = Re(z)+Re(w)$ for any complex $z,w$.
$Im(z+w) = Im(z)+Im(w)$ for any complex $z,w$.
You may first identify real part and imaginary part:
$$x + yi + \sqrt{x^2 +y^2} = 8 + 4i\implies x+ \sqrt{x^2 +y^2}=8, yi=4i $$
$$ y=4, x+ \sqrt{x^2 +16}=8 $$
$$x=3, y=4 \implies z=3+4i$$