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