How can I get the real number and the imaginary number from: $$\frac{3+i}{5-12i}$$
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$$\dfrac{a+ib}{c+id}=\dfrac{(a+ib)(c-id)}{(c+id)(c-id)}=\dfrac{ac+bd+i(bc-ad)}{c^2+d^2}$$
lab bhattacharjee
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I reached $27+41i/169$ and i'm not sure how canIi get the real and imaginary number from it – D_R May 13 '15 at 16:34
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1@DinRevah, $$\dfrac{x+iy}n=\dfrac xn+i\dfrac yn$$ for $x,y,n$ are real numbers – lab bhattacharjee May 13 '15 at 16:37
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$$\frac{3+i}{5-12i}=$$
$$\left|\frac{3+i}{5-12i}\right|e^{\arg\left(\frac{3+i}{5-12i}\right)i}=$$
$$\frac{\left|3+i\right|}{\left|5-12i\right|}e^{\arg\left(\frac{3}{169}+\frac{41}{169}i\right)i}=$$
$$\frac{\sqrt{3^2+1^2}}{\sqrt{5^2+12^2}}e^{\tan^{-1}\left(\frac{41}{3}\right)i}=$$
$$\frac{\sqrt{10}}{\sqrt{169}}e^{\tan^{-1}\left(\frac{41}{3}\right)i}=$$
$$\frac{\sqrt{10}}{13}e^{\tan^{-1}\left(\frac{41}{3}\right)i}=$$
$$\frac{\sqrt{10}}{13}\left(\cos\left(\tan^{-1}\left(\frac{41}{3}\right)\right)+\sin\left(\tan^{-1}\left(\frac{41}{3}\right)\right)i\right)=$$
$$\frac{\sqrt{10}}{13}\cos\left(\tan^{-1}\left(\frac{41}{3}\right)\right)+\frac{\sqrt{10}}{13}\sin\left(\tan^{-1}\left(\frac{41}{3}\right)\right)i=$$
$$\frac{\sqrt{10}}{13}*\frac{3}{13\sqrt{10}}+\frac{\sqrt{10}}{13}*\frac{41}{13\sqrt{10}}i$$
$$\frac{3}{169}+\frac{41}{169}i$$
Jan Eerland
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This is the long answer, but if you want to understand it, this the way to calculate it! – Jan Eerland May 13 '15 at 16:45
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That seems like overkill, just to show that $3+\frac i5-12i=3-\frac{59}5i$! Besides, turns out he meant $\frac{3+i}{5-12i}$, anyway. – Akiva Weinberger May 13 '15 at 17:39
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