Solving $|z|i+2z=\sqrt{3}$

Question

How one can solve the following complex equation, where $z$ is complex number.

$$|z|i+2z=\sqrt{3}$$

Thank you.

Answer 1

HINT:

$$|z|=\frac{\sqrt3-2z}i=2zi-\sqrt3i=i(2z-\sqrt3)$$

If $z=a+ib,2z-\sqrt3=2a-\sqrt3+ib\iff i(2z-\sqrt3)=-b+i(2a-\sqrt3)$

$\displaystyle\implies 2a-\sqrt3=0$

Answer 2

See $z$ as $a+ib$ and $|z|=\sqrt{a^2+b^2}$ you need nothing more than that...

$|z|i+2z=\sqrt{3}\Rightarrow i\sqrt{a^2+b^2}+2(a+ib)=\sqrt{3}\Rightarrow 2a+i(**)=\sqrt{3}$

Just compare real/imaginary part both sides....

Answer 3

$$|z|i+2z=\sqrt{3}$$

which is also: $$ \sqrt {a^2+b^2}i+2a+2bi=\sqrt{3} +0i$$ compare real and imaginary part concordantly;

You'll get the system:

$ \sqrt {a^2+b^2} +2b= 0$

and

$ 2a= \sqrt{3}$

Solve for $a$ and $ b$, thus you'll find the requested $z$.

Answer 4

Decomposing into real part and imaginary part is really not needed...

For every solution $z$, $2z=\sqrt3-\mathrm i|z|$ $(\ast)$, in particular the modules must coincide hence $|2z|^2=(\sqrt3)^2+(|z|)^2$, that is, $4|z|^2=3+|z|^2$, or simply, $|z|=1$. Using $(\ast)$ once again, one sees that $2z=\sqrt3-\mathrm i$, hence the value of $z$. Now check that this value of $z$ solves the equation and you are done.