# Help complex numbers

• Jul 6th 2006, 03:38 PM
Brooke
Help 1 ???
Perform the indicated operations: (2+5i)(1-3i) all over 5-2i
• Jul 6th 2006, 04:03 PM
Soroban
Hello, Brooke!

I assume you can do basic Algebra and that you know that $\displaystyle i^2 = -1$

Quote:

Perform the indicated operations: $\displaystyle \frac{(2+5i)(1-3i)}{5-2i}$

Simplify the numerator: .$\displaystyle (2 + 5i)(1 - 3i) \:=\:2 - 6i + 5i - 15i^2$

. . and we have: .$\displaystyle 2 - i -15(-1)\:=\:2 - i + 15 \:= \:17 - i$

The problem becomes: .$\displaystyle \frac{17 - i}{5 - 2i}$

Now we are expected to rationalize the denominator.

Multiply top and bottom by the conjugate of the denominator:

. . $\displaystyle \frac{17 - i}{5 - 2i}\cdot\frac{5 + 2i}{5 + 2i} \;= \;\frac{85 + 34i - 5i - 2i^2}{25 + 10i - 10i - 4u^2} \;=$ $\displaystyle \frac{85 + 29i - 2(-1)}{25 - 4(-1)}$

. . $\displaystyle = \;\frac{85 + 29i + 2}{25 + 4}\;=\;\frac{87+29i}{29} \;=$ $\displaystyle \frac{\not{29}(3 + i)}{\not{29}} \;= \;3 + i$