# How do you find z such that (4−3i) /z+(−5+2i) = (−2−2i) ?

##### 1 Answer
Jun 19, 2016

$z = \frac{24}{25} + \frac{7}{25} i$

#### Explanation:

Write as:$\text{ "" } \frac{4 - 3 i}{z} = \left(- 2 - 2 i\right) - \left(- 5 + 2 i\right)$

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

$\frac{4 - 3 i}{z} = \frac{3 - 4 i}{1}$

Turn everything upside down:

$\frac{z}{4 - 3 i} = \frac{1}{3 - 4 i}$

$z = \frac{4 - 3 i}{3 - 4 i}$

Multiply by $1$ but in the form of $1 = \frac{3 + 4 i}{3 + 4 i}$, which is the complex conjugate of the denominator.

$z = \frac{4 - 3 i}{3 - 4 i} \times \frac{3 + 4 i}{3 + 4 i} = \frac{\left(4 - 3 i\right) \left(3 + 4 i\right)}{\left(3 - 4 i\right) \left(3 + 4 i\right)}$

Distribute:

$z = \frac{12 + 16 i - 9 i - 12 {i}^{2}}{9 + 12 i - 12 i - 16 {i}^{2}} = \frac{12 + 7 i - 12 {i}^{2}}{9 - 16 {i}^{2}}$

Since $i = \sqrt{- 1}$, we know that ${i}^{2} = - 1$:

$z = \frac{12 + 7 i - 12 \left(- 1\right)}{9 - 16 \left(- 1\right)} = \frac{12 + 7 i + 12}{9 + 16} = \frac{24 + 7 i}{25}$

$z = \frac{24}{25} + \frac{7}{25} i$