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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.

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  • $\begingroup$ Maybe it is $16+15 i$ ? $\endgroup$ – Emilio Novati Dec 3 '17 at 22:16
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    $\begingroup$ Hint: Compare real and imaginary parts. $\endgroup$ – John Doe Dec 3 '17 at 22:19
  • $\begingroup$ rewriting the left-hand side as a sum real and imaginary you get $m=-3$, $n=7$. $\endgroup$ – daulomb Dec 3 '17 at 22:21
  • $\begingroup$ 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. $\endgroup$ – Rebellos Dec 3 '17 at 22:22
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    $\begingroup$ Yes, I did mean $16 + 15i$, thank you for the edit. $\endgroup$ – user509838 Dec 3 '17 at 22:23
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$$(4-5i)m + 4n = 16+15i \Leftrightarrow 4m - 5im + 4n = 16 + 15i \Leftrightarrow 4(m+n)+i(-5m)=16+15i$$

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 ?

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

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Hint:

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

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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 $n$.

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