Question #8380d

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Question #8380d

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Answer 1 · 2018-02-27T06:56:51

Did you mean to put the negative sign after the divide sign?
If so, $\frac{4 - 3 i + 2 + 6 i}{- ((3 - i) (3 - i))} = - \frac{3}{10} - \frac{3}{5} i$ $(4-3i+2+6i)/-((3-i)(3-i))=-3/10-3/5i$ .

Explanation:

First, simplify the numerator by combining the integer terms and combining the $i$ $i$ terms: $4 - 3 i + 2 + 6 i = 3 i + 6$ $4-3i+2+6i=3i+6$ .

Secondly, the denominator: $- ((3 - i) (3 - i)) = - (9 - 3 i - 3 i + i^{2})$ $-((3-i)(3-i))=-(9-3i-3i+i^2)$ . Since $i = \sqrt{- 1}$ $i=sqrt(-1)$ , $i^{2} = {(\sqrt{- 1})}^{2} = - 1$ $i^2=(sqrt(-1))^2=-1$ .
Plugging that into the denominator and combining like terms, we get $- (9 - 6 i + (- 1)) = - (8 - 6 i) = - 8 + 6 i = 6 i - 8$ $-(9-6i+(-1))=-(8-6i)=-8+6i=6i-8$

Our fraction is now $\frac{3 i + 6}{6 i - 8}$ $(3i+6)/(6i-8)$

This may be okay for your teacher, but the standard form for complex numbers is $a \pm b i$ $a+-bi$ , so let's get our answer in that format by multiplying the numerator and denominator by the conjugate of the denominator: $\frac{(3 i + 6) (6 i + 8)}{(6 i - 8) (6 i + 8)}$ $((3i+6)(6i+8))/((6i-8)(6i+8))$
We get $\frac{18 i^{2} + 24 i + 36 i + 48}{- 36 - 64} = \frac{30 + 60 i}{- 100}$ $(18i^2+24i+36i+48)/(-36-64)=(30+60i)/-100$ and put it in the form of $a \pm b i$ $a+-bi$ then reduce it:
$\frac{30}{- 100} - \frac{60 i}{- 100} = - \frac{3}{10} - \frac{3}{5} i$ $30/-100-(60i)/-100=-3/10-3/5i$

Hope that helps! Have fun; imaginary numbers are cool!

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Question #8380d

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