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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!

  • 3
    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.2012-10-07

2 Answers 2

6

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

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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$