How do you divide (4+5i)/(2-3i)?

Jan 10, 2016

Multiply both numerator and denominator by the Complex conjugate of the denominator, then simplify to find:

$\frac{4 + 5 i}{2 - 3 i} = - \frac{7}{13} + \frac{22}{13} i$

Explanation:

Multiply numerator and denominator by $2 + 3 i$ (using FOIL if it helps) and simplify:

$\frac{4 + 5 i}{2 - 3 i}$

$= \frac{\left(4 + 5 i\right) \left(2 + 3 i\right)}{\left(2 - 3 i\right) \left(2 + 3 i\right)}$

$= \frac{\stackrel{\text{First" overbrace((4*2))+stackrel "Outside" overbrace((4*3i))+stackrel "Inside" overbrace((5i*2))+stackrel "Last}}{\overbrace{\left(5 i \cdot 3 i\right)}}}{{2}^{2} + {3}^{2}}$

$= \frac{8 + 12 i + 10 i - 15}{4 + 9}$

$= \frac{- 7 + 22 i}{13}$

$= - \frac{7}{13} + \frac{22}{13} i$

Jan 10, 2016

$= - \frac{7}{13} + \frac{22}{13} i$

Explanation:

$\frac{4 + 5 i}{2 - 3 i}$

Whenever we divide complex numbers we multiply both numerator and denominator with the complex conjugate of the denominator, this makes the denominator a real number

If the complex number is $a + i b$ then the complex conjugate is $a - i b$

For example
$\left(a + i b\right) \left(a - i b\right)$
$= {\left(a\right)}^{2} - {\left(i b\right)}^{2}$
$= {a}^{2} - {i}^{2} {b}^{2}$
$= {a}^{2} - \left(- 1\right) {b}^{2}$
$= {a}^{2} + {b}^{2}$ this is a real number.

Now back to our problem.

$\frac{4 + 5 i}{2 - 3 i}$

$= \frac{4 + 5 i}{2 - 3 i} \cdot \frac{2 + 3 i}{2 + 3 i}$

$= \frac{\left(4 + 5 i\right) \left(2 + 3 i\right)}{{2}^{2} + {3}^{2}}$

$= \frac{4 \left(2\right) + 4 \left(3 i\right) + 5 i \left(2\right) + 5 i \left(3 i\right)}{4 + 9}$

$= \frac{8 + 12 i + 10 i + 15 {i}^{2}}{13}$

$= \frac{8 + 22 i + 15 \left(- 1\right)}{13}$

$= \frac{8 - 15 + 22 i}{13}$

$= \frac{- 7 + 22 i}{13}$ Answer

$= - \frac{7}{13} + \frac{22}{13} i \quad$ Answer in $a + i b$ form