Help complex numbers

**Brooke** · July 6th 2006, 03:38 PM

Perform the indicated operations: (2+5i)(1-3i) all over 5-2i

**Soroban** · July 6th 2006, 04:03 PM

Hello, Brooke!

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

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

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

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

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

Now we are expected to rationalize the denominator.

Multiply top and bottom by the conjugate of the denominator:

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

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

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