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I have found a particular solution to a second order differential equation to be: $$\left(\frac{10}{(1+i)}\right)e^{ix}$$ However, the solution is not the problem, my problem is that I have used Wolfram alpha to simplify the coefficient and found that: $$\left(\frac{10}{(1+i)}\right)=5-5i.$$ I do not understand how they are equal. I vaguely remember some identity regarding -i, which I'm assuming is how I simplify, if someone could let me know how I get from LHS to RHS that'd be great. Thanks

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    $\begingroup$ Just multiply by the conjugate of $1 + i$ both the numerator and denominator. $\endgroup$
    – Azlif
    Jan 7, 2018 at 14:17

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Because $$\frac{10}{1+i}=\frac{10(1-i)}{(1+i)(1-i)}=\frac{10(1-i)}{2}=5(1-i).$$

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If you do no remember that you have to multiply by the conjugated quantity (the preferred method anyway, since it works also with square roots), you can always try direct identification:

$\dfrac{10}{1+i}=a+ib\iff 10=(a+ib)(1+i)=(a-b)+i(a+b)$

So you solve the system $\begin{cases}a-b=10\\a+b=0\end{cases}\iff\begin{cases}a=5\\b=-5\end{cases}\quad$ and find $5-5i$.

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