# How do you solve z( 3 + 2i) = (2 — 3i)?

Dec 29, 2015

$z = i$

#### Explanation:

$z \left(3 + 2 i\right) = \left(2 - 3 i\right)$

To solve for $z$ divide both sides by $3 + 2 i$ like this

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

$\implies z = \frac{2 - 3 i}{3 + 2 i}$

Then multiply by the conjugate of the denominator to simplify the expression

z= ((2-3i)/(3+2i))*color(red)(((3-2i)/(3-2i))

Then multiply and combine like terms like this

$z = \frac{6 - 4 i - 9 i + 6 {i}^{2}}{9 - 6 i + 6 i - 4 {i}^{2}}$

$\implies \frac{6 - 13 i - + 6 {i}^{2}}{9 - 4 {i}^{2}}$

Replace ${i}^{2} = - 1$ then simplify

$\implies z = \frac{6 - 13 i + 6 \cdot \textcolor{b l u e}{- 1}}{9 - 4 \cdot \textcolor{b l u e}{- 1}}$

$\implies z = \frac{6 - 6 - 13 i}{9 + 4}$

$\implies z = \frac{- 13 i}{13} = i$

$z = i$

I hope this helps :)