How do you simplify $7 + 4 i \div 2 - 3 i$ $7+4i div 2-3i$ ?

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How do you simplify $7 + 4 i \div 2 - 3 i$ $7+4i div 2-3i$ ?

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Answer 1 · 2016-01-12T16:48:40

$\frac{2}{13} + \frac{29}{13} i$ $2/13 + 29/13 i$

Explanation:

Basically, $(7 + 4 i) \div (2 - 3 i)$ $(7 + 4i) -: (2 - 3i)$ is the same as the fraction $\frac{7 + 4 i}{2 - 3 i}$ $( 7 + 4i ) / (2 - 3i)$ . As I prefer to work with fractions, I will stick with this formulation.

To simplify $\frac{7 + 4 i}{2 - 3 i}$ $( 7 + 4i ) / (2 - 3i)$ , you need to find the complex conjugate of the denominator and extend the fraction with it:

Your denominator is $2 - 3 i$ $2 - 3i$ , so the complex conjugate is $2 + 3 i$ $2 + 3i$ .

You need to extend the fraction with $2 + 3 i$ $2 + 3i$ , i.e. multiply both the numerator and the denominator by it:

$\frac{7 + 4 i}{2 - 3 i} = \frac{(7 + 4 i) \cdot (2 + 3 i)}{(2 - 3 i) \cdot (2 + 3 i)} = \frac{14 + 8 i + 21 i + 12 i^{2}}{2^{2} - {(3 i)}^{2}} = \frac{14 + 29 i + 12 i^{2}}{4 - 9 i^{2}}$ $( 7 + 4i ) / (2 - 3i) = (( 7 + 4i )* (2 + 3i)) / ((2 - 3i)*(2 + 3i)) = (14 + 8i + 21i + 12 i^2) / (2^2 - (3i)^2) = (14 + 29i + 12 i^2) / (4 - 9 i^2)$

... remember that $i^{2} = - 1$ $i^2 = -1$ ...

$= \frac{14 - 12 + 29 i}{4 + 9} = \frac{2 + 29 i}{13} = \frac{2}{13} + \frac{29}{13} i$ $= (14 - 12 + 29i ) / (4 + 9) = (2 + 29i) / 13 = 2/13 + 29/13 i$

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How do you simplify $7 + 4 i \div 2 - 3 i$ $7+4i div 2-3i$ ?

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