Test 4 Section 3 #14

When you’re dividing complex numbers, you have to multiply the top and bottom of the fraction by the complex conjugate of the bottom. This creates a real number in the bottom of the fraction, which is awesome.

The bottom of this fraction is 3-2i, so its complex conjugate is 3+2i. To get started, then, we write the following:

    \begin{align*}\dfrac{8-i}{3-2i}\times\dfrac{3+2i}{3+2i}\end{align*}

Then we simplify as much as we can. First we FOIL:

    \begin{align*}&\dfrac{(8-i)(3+2i)}{(3-2i)(3+2i)}\\=&\dfrac{24+16i-3i-2i^2}{9+6i-6i-4i^2}\\=&\dfrac{24+13i-2i^2}{9-4i^2}\end{align*}

Now, remembering that i=\sqrt{-1} and therfore that i^2=-1, we make that substitution and simplify further:

    \begin{align*}&\dfrac{24+13i-2(-1)}{9-4(-1)}\\=&\dfrac{24+13i+2}{9+4}\\=&\dfrac{26+13i}{13}\\=&2+i\end{align*}

So there you have it. Once we simplify that complex fraction into a+bi form, we see that it’s just 2+i, which means a=2.

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