# How to solve a complex linear equation with two variables

In a review question I get the equation $(4-5i)m + 4n = 16+15i$. Where $i$ is the imaginary unit, $m$ and $n$ are real numbers. I do not know how to go about solving this equation. There is also another section to the question which asks to solve it when $m$ and $n$ are conjugate complex numbers.

Thank you.

• Maybe it is $16+15 i$ ? – Emilio Novati Dec 3 '17 at 22:16
• Hint: Compare real and imaginary parts. – John Doe Dec 3 '17 at 22:19
• rewriting the left-hand side as a sum real and imaginary you get $m=-3$, $n=7$. – daulomb Dec 3 '17 at 22:21
• I assume you meant $16+15i$ as @EmilioNovati mentioned, so I gave it a heads-up correction. Re-edit it if that's not the case. – Rebellos Dec 3 '17 at 22:22
• Yes, I did mean $16 + 15i$, thank you for the edit. – user509838 Dec 3 '17 at 22:23

$$(4-5i)m + 4n = 16+15i \Leftrightarrow 4m - 5im + 4n = 16 + 15i \Leftrightarrow 4(m+n)+i(-5m)=16+15i$$

So, by the identity of complex numbers, we'll get :

$$\begin{cases} 4(m+n)=16 \\ -5m = 15\end{cases}$$

can you solve that system of linear equations now and yield your solution ?

• @JohnDoe I should go sleep as it seems :D – Rebellos Dec 3 '17 at 22:26
• Haha yes, you got through most of it correctly at least! :) I will delete the above comment, and this one too shortly. – John Doe Dec 3 '17 at 22:27
• This is a great explanation, I got the solutions $m = -3$ and $n = 7$. Thank you. – user509838 Dec 3 '17 at 22:31

For $m$ and $n$ real, just compare real and imaginary parts.

For $m$ and $n$ complex numbers conjugate to one another, just substitute $a + bi$ for $n$ and $a - bi$ for $m$, where $a$ and $b$ are real numbers. To finish just again compare real and imaginary parts.

Hint:

use the identity of complex numbers: $$a+ib=m+in \quad \iff \quad a=m \quad \mbox{and} \quad b=m$$

The equation you end up with is $$(4-5i)m+4n=16+15i\\(4m+4n)-(5m)i=16+15i$$ Compare imaginary parts to get $m$, then real parts to get $n$.

When $m$ and $n$ are complex conjugates, write $m=a+bi$, $n=a-bi$. Then substitute this into the above equation, multiply out and the compare real and imaginary parts again to obtain $a$ and $b$, thus getting $m$ and $n$.