Question

How do I determine z ∈ C:

Answer 1

Simplify the coefficient of z on the left side. We do this by rationalizing the denominators and multiplying them by their complex conjugates:

$\dfrac{3-2i}{1+i} - \dfrac{5+3i}{1+2i} = \dfrac{3-2i}{1+i}\cdot\dfrac{1-i}{1-i} - \dfrac{5+3i}{1+2i}\cdot\dfrac{1-2i}{1-2i}$

$\dfrac{3-2i}{1+i} - \dfrac{5+3i}{1+2i} = \dfrac{(3-2i)(1-i)}{1-i^2} - \dfrac{(5+3i)(1-2i)}{1-(2i)^2}$

$\dfrac{3-2i}{1+i} - \dfrac{5+3i}{1+2i} = \dfrac{3 - 2i - 3i + 2i^2}{1-(-1)} - \dfrac{5 + 3i - 10i - 6i^2}{1-4(-1)}$

$\dfrac{3-2i}{1+i} - \dfrac{5+3i}{1+2i} = \dfrac{3 - 5i + 2(-1)}2 - \dfrac{5 - 7i - 6(-1)}5$

$\dfrac{3-2i}{1+i} - \dfrac{5+3i}{1+2i} = \dfrac{1 - 5i}2 - \dfrac{11 - 7i}5$

$\dfrac{3-2i}{1+i} - \dfrac{5+3i}{1+2i} = \dfrac{1 - 5i}2\cdot\dfrac55 - \dfrac{11 - 7i}5\cdot\dfrac22$

$\dfrac{3-2i}{1+i} - \dfrac{5+3i}{1+2i} = \dfrac{5 - 25i - 22 + 14i}{10}$

$\dfrac{3-2i}{1+i} - \dfrac{5+3i}{1+2i} = -\dfrac{17 + 11i}{10}$

So, the equation is simplified to

$-\dfrac{17+11i}{10} z = \dfrac12 - \dfrac{2i}5$

Let's combine the fractions on the right side:

$\dfrac12 - \dfrac{2i}5 = \dfrac12\cdot\dfrac55 - \dfrac{2i}5\cdot\dfrac22$

$\dfrac12 - \dfrac{2i}5 = \dfrac{5-4i}{10}$

Then

$-\dfrac{17+11i}{10} z = \dfrac{5-4i}{10}$

reduces to

$-(17+11i) z = 5-4i$

Multiply both sides by -1/(17 + 11i) :

$\dfrac{-(17+11i)}{-(17+11i)} z = \dfrac{5-4i}{-(17+11i)}$

$z = -\dfrac{5-4i}{17+11i}$

Finally, simplify the right side:

$-\dfrac{5-4i}{17+11i} = -\dfrac{5-4i}{17+11i} \cdot \dfrac{17-11i}{17-11i}$

$-\dfrac{5-4i}{17+11i} = -\dfrac{(5-4i)(17-11i)}{17^2-(11i)^2}$

$-\dfrac{5-4i}{17+11i} = -\dfrac{85 - 68i - 55i + 44i^2}{289-121(-1)}$

$-\dfrac{5-4i}{17+11i} = -\dfrac{85 - 68i - 55i + 44(-1)}{410}$

$-\dfrac{5-4i}{17+11i} = -\dfrac{41 - 123i}{410}$

$-\dfrac{5-4i}{17+11i} = -\dfrac{41 - 41\cdot3i}{410}$

$-\dfrac{5-4i}{17+11i} = -\dfrac{1 - 3i}{10}$

So, the solution to the equation is

$z = -\dfrac{1-3i}{10} = \boxed{-\dfrac1{10} + \dfrac3{10}i}$