How do you simplify 1/(2+3i) and write in a+bi form?

Dec 5, 2015

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

Explanation:

Multiply by the complex conjugate.

$\frac{1}{2 + 3 i} \left(\frac{2 - 3 i}{2 - 3 i}\right) = \frac{2 - 3 i}{4 - 6 i + 6 i - 9 {i}^{2}} = \frac{2 - 3 i}{4 - 9 {i}^{2}}$

Recall that ${i}^{2} = - 1$ since $i = \sqrt{- 1}$.

$= \frac{2 - 3 i}{4 - \left(- 9\right)} = \frac{2 - 3 i}{13} = \frac{2}{13} - \frac{3}{13} i$

Dec 5, 2015

$= 0 , 1538 - 0 , 23078 i$

Explanation:

$1$ may be written in rectangular form as $1 + 0 i$, and in polar form as $1 \angle {0}^{\circ}$.

Similarly, $2 + 3 i$ can be written in polar form as
$\sqrt{{2}^{2} + {3}^{2}} \angle {\tan}^{- 1} \left(\frac{3}{2}\right) = \sqrt{13} \angle 56 , {31}^{\circ}$

We may now divide them by using the standard rule for division of complex numbers in polar form :

$\frac{1 \angle 0}{\sqrt{13} \angle 56 , {31}^{\circ}} = \frac{1}{\sqrt{13}} \angle \left(0 - 56 , 31\right)$

This answer may now be converted back to rectangular form using standard trig relationships to get $\left(\frac{1}{\sqrt{13}}\right) \cos \left(- 56 , 31\right) + i \left(\frac{1}{\sqrt{13}}\right) \sin \left(- 56 , 31\right)$

$= 0 , 1538 - 0 , 23078 i$

(Note : We could have maintained complex rectangular form and multiply top and bottom by the complex conjugate of the denominator $\overline{z} = 2 - 3 i$ and then multiply top and bottom. Bottom would then be a real number, and top would require rule of multiplication of complex numbers in rectangular form - quite a complex rule ;) - Hence probably quicker the way I did it by rather converting everything to polar form first)