(3+2i)/(1-2i)=z equation
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The solution
Detail solution
Given the linear equation:
(3+2*i)/(1-2*i) = z
Expand brackets in the left part
3+2*i1-2*i = z
Looking for similar summands in the left part:
(1 + 2*i)*(3 + 2*i)/5 = z
Move the summands with the unknown z
from the right part to the left part:
$$- z + \frac{\left(1 + 2 i\right) \left(3 + 2 i\right)}{5} = 0$$
Divide both parts of the equation by (-z + (1 + 2*i)*(3 + 2*i)/5)/z
z = 0 / ((-z + (1 + 2*i)*(3 + 2*i)/5)/z)
We get the answer: z = -1/5 + 8*i/5
$$z_{1} = - \frac{1}{5} + \frac{8 i}{5}$$
Sum and product of roots
[src]
$$- \frac{1}{5} + \frac{8 i}{5}$$
$$- \frac{1}{5} + \frac{8 i}{5}$$
$$- \frac{1}{5} + \frac{8 i}{5}$$
$$- \frac{1}{5} + \frac{8 i}{5}$$