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This was the problem I was given:

Compute the complex number for $\frac{(18-7i)}{(12-5i)}$.

I was told to write this in the form of $a+bi$.

So please give me a hint of how to do this. :)

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closed as off-topic by Daniel Robert-Nicoud, Daniel W. Farlow, John B, MCT, T. Bongers Apr 27 '16 at 4:38

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    $\begingroup$ Multiply the numerator and denominator by $12+5i$. $\endgroup$ – Edward Jiang Apr 24 '16 at 18:50
  • $\begingroup$ Perhaps you want to write it in the form $a+bi$ (not $ai$). Use the conjugate of the denominator. $\endgroup$ – Michael Burr Apr 24 '16 at 18:51
  • $\begingroup$ Use identity $$\frac1z=\frac{\overline z}{\lvert z\rvert^2}$$or, if you are more familiar with it $$a,b\in\Bbb R,\quad \frac{1}{a+bi}=\frac{a-bi}{a^2+b^2}=\frac{a}{a^2+b^2}-\frac{b}{a^2+b^2}i$$ $\endgroup$ – user228113 Apr 24 '16 at 18:52
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Proceeding as in my comment: $$\frac{(18-7i)(12+5i)}{(12-5i)(12+5i)}$$ $$=\frac{216+90i-84i+35}{144++60i-60i+25}$$ $$=\frac{251+6i}{169}$$ $$=\frac{251}{169}+\frac{6i}{169}$$

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    $\begingroup$ @malcolm Yes, that's the definition. $\endgroup$ – Edward Jiang Apr 24 '16 at 18:59
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Use the complex conjugate of the denominator:

\begin{equation} \frac{18-7i}{12-5i} = \frac{18-7i}{12-5i}\frac{12+5i}{12+5i} = \frac{1}{12^2 + 5^2}[216 + 35 - 84i + 90i] = \frac{251}{169} + \frac{6}{169}i \end{equation}

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You can "rationalize" or more accurately "real-ize" the denominator by multiplying the numerator and denominator by denominator's conjugate. It is just like with radicals. You will get:

$$\frac{18-7i}{12-5i}=\frac{(18-7i)(12+5i)}{(12-5i)(12+5i)}=\frac{216-84i+90i-35i^2}{144-60i+60i-25i^2}=\frac{216+6i+35}{144+25}=\frac{251+6i}{169}$$

More accurate answer would be: $$\frac{251+6i}{169}=\frac{251}{169}+\frac{6}{169}i$$

If you want the decimal answer, it is around $1.485207101+0.035502959i$.

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