# How to solve complex equation with same variable on two sides

What is the analytic solution of X for the equation below?

$$conjugate(X)= \frac{-2\times A}{B\times X}$$

A, B and X are complex numbers.

Would the magnitude of X be given by this? $$|X| =\sqrt \frac {-2\times A}{B}$$

Then what would be the solution for the angle? Can I get a polynomial solution for X_real and X_imag without having to use atan(.) function?

• I'm assuming by "$conjugate(X)$" you mean e.g. $a - bi$ if $X = a + bi$? Commented Sep 18, 2015 at 11:40
• Math1000, yes, that is correct.
– Joy
Commented Sep 18, 2015 at 11:41
• Lee Yiyuan, I am unsure that it gives me the complete solution. I wonder how to get the angle. Also, this places constraints on values of A & B I guess.
– Joy
Commented Sep 18, 2015 at 11:43

You want to solve $$\bar z=-\frac{2a}{bz}$$ for $z$, with $z,a,b\in\Bbb C$.

First of all $b,z\neq0$.

Multipliyng both sides by $z$, you get $|z|^2=z\bar z=-\frac{2a}{b}$ thus your thoughts about the magnitude were correct.

Then to solve, just write these complex numbers in their polar form: $$a=re^{i\theta}\\ b=se^{i\psi}\\ z=te^{i\vartheta} \\ \mbox{(in particular being z,b\ne0 you have t,s>0)}$$ thus you get $$t^2=-2\frac rse^{i(\theta-\psi)}$$ from which you deduce that $e^{i(\theta-\psi)}$ must be real and negative: thus $\theta-\psi\in\pi+2\pi\Bbb Z$, thus the previous equation turns into $$t^2=2r/s$$ so our solutions are all and only $te^{i\vartheta}$ with $t=\sqrt{2r/s}$ and $\vartheta\in\Bbb R$ (i.e. every angle solves the equation).

Then if we want to write the solutions in an "analytic" form, just pass from polar to algebraic form: $$z=\sqrt{2\frac rs}e^{i\vartheta}=\sqrt{2\frac rs}\cos\vartheta+i\sqrt{2\frac rs}\sin\vartheta\;\;.$$

• Thank you very much Joe. Now I am clear that the condition θ−ψ∈π+2πZ will arise. The angle can be anything. Can we then say we have the analytical expression for z_real & z_imag? Or do we have to find a mechanism to choose the angle?
– Joy
Commented Sep 18, 2015 at 11:58
• I was wondering how I can represent $X_{real}$ and $X_{imag}$ using analytical expressions.
– Joy
Commented Sep 18, 2015 at 12:04
• I edited... did I answerd you?
– Joe
Commented Sep 18, 2015 at 13:15
• I had initially expected an analytical expression where RHS contained only the known constants from the equation. I now see that the solution to the original equation is a circle and not a point. And the expression will have to contain the free-variable representing the angle of z. So, I think you have already answered. My initial expectation of having $X_{real}$ and $X_{imag}$ only in terms of the constants $A_{real}, A_{imag}, B_{real}, B_{imag}$ seems incorrect.
– Joy
Commented Sep 18, 2015 at 15:23

Because the two equations you wrote are equivalent if $X \neq 0$, every angle gives you a valid solution. You have $$\bar X= \frac{-2\times A}{B\times X} \Leftrightarrow |X|^2 = X \bar X = \frac{-2A}B$$

There are only solutions, if $\frac{-2 a}B > 0$

• Correct me if I am wrong, but I would add that there are only solutions of $\frac{-2 a}{b}\in\mathbb{R}^+$. Commented Sep 18, 2015 at 11:46
• Supinf, only the first equation is the one to be solved. Second equation seems correct expression for the magnitude, but also seems as if I am limiting the possible answers. It looks like $\frac {-2A}{B}$ must be REAL! Yes, the magnitude must be certainly non-negative.
– Joy
Commented Sep 18, 2015 at 11:49
• @Joy the second and the first equation are equivalent, so you can solve the second equation instead of the first (but you have to take extra care for the case that $X=0$ Commented Sep 18, 2015 at 11:53
• Thank you supinf. That is correct. I was wondering how I can represent $X_{real}$ and $X_{imag}$ using analytical expressions.
– Joy
Commented Sep 18, 2015 at 12:01
• in that case you would have the equation $X_{real}^2 + X_{imag}^2 = \frac{-2A}B$ - which gives you the same solutions Commented Sep 18, 2015 at 12:04