How do you simplify $\frac{3 i}{9 - 6 i}$ $(3i)/(9-6i)$ ?

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How do you simplify $\frac{3 i}{9 - 6 i}$ $(3i)/(9-6i)$ ?

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Answer 1 · 2015-12-31T07:40:01

$- \frac{2}{13} + \frac{3}{13} i$ $-2/13+3/13i$

Explanation:

Divide each term by $3$ $3$ .

$\Rightarrow \frac{i}{3 - 2 i}$ $=>i/(3-2i)$

Now, to remove the imaginary part from the denominator and write the answer as a complex number in the form $a + b i$ $a+bi$ , multiply the fraction by the complex conjugate of the denominator.

$\Rightarrow \frac{i}{3 - 2 i} (\frac{3 + 2 i}{3 + 2 i})$ $=>i/(3-2i)((3+2i)/(3+2i))$

Distribute. Notice that the bottom will form a difference of squares.

$\Rightarrow \frac{3 i + 2 i^{2}}{9 - 4 i^{2}}$ $=>(3i+2i^2)/(9-4i^2)$

To continue simplifying, rewrite $i^{2}$ $i^2$ as $- 1$ $-1$ . $i^{2} = - 1$ $i^2=-1$ since $i = \sqrt{- 1}$ $i=sqrt(-1)$ .

$\Rightarrow \frac{3 i + 2 (- 1)}{9 - 4 (- 1)} = \frac{- 2 + 3 i}{13}$ $=>(3i+2(-1))/(9-4(-1))=(-2+3i)/13$

Split apart the numerator to write in $a + b i$ $a+bi$ form.

$\Rightarrow - \frac{2}{13} + \frac{3}{13} i$ $=>-2/13+3/13i$

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How do you simplify $\frac{3 i}{9 - 6 i}$ $(3i)/(9-6i)$ ?

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How do you simplify $\frac{3 i}{9 - 6 i}$ $(3i)/(9-6i)$ ?