2
$\begingroup$

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
    Hint: Try multiplying by $(9-3i)/(9-3i)$.2012-10-07
  • 1
    You might want to extract obvious factors to give $z = \cfrac 5 3 .\cfrac 1 {3+i} .\cfrac {3-i}{3-i}$2012-10-07
  • 2
    Going with Sean's hint, think of what happens when you multiply a complex number by its conjugate.2012-10-07
  • 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