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Into its real and imaginary components? Wolfram tells me it's equivalent to $\frac{1}{2}+\frac{i}{6}$, but I don't know how to arrive there myself.

Thank you!

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3  
Hint: Try multiplying by $(9-3i)/(9-3i)$. –  Sean Eberhard Oct 7 '12 at 20:47
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You might want to extract obvious factors to give $z = \cfrac 5 3 .\cfrac 1 {3+i} .\cfrac {3-i}{3-i}$ –  Mark Bennet Oct 7 '12 at 20:50
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Going with Sean's hint, think of what happens when you multiply a complex number by its conjugate. –  Domonic Mei Oct 7 '12 at 20:51
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The general procedure is to multiply the numerator and denominator of the given complex fraction by the complex conjugate of the denomnator when that denominator is not real. –  Geoff Robinson Oct 7 '12 at 21:02

2 Answers 2

$z = \frac{5}{9+3i} = \frac{5(9-3i)}{(9+3i)(9-3i)} = \frac{45-15i}{9^2 + 3^2} = \frac{45}{90} - \frac{15i}{90} = \frac{1}{2} - \frac{i}{6}$

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Here, we must use our knowledge of complex numbers and their conjugates as well as our knowledge of the difference of squares:

$z = \frac{5}{9+3i}$

$ = \frac{5(9-3i)}{(9+3i)(9-3i)}$

$ = \frac{45-15i}{9^2 - (i^2)(3^2)}$

$ = \frac{45-15i}{90}$

$ = \frac{1}{2} - \frac{1}{6}i$

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