My Coding > Mathematics > Complex numbers > Solve z^2 + 6iz - 5z - 9i - 1 = 0

Solve z^2 + 6iz - 5z - 9i - 1 = 0

In this short lecture, I will show you how to solve a classical square equation with complex numbers instead of real ones. For this example, I will solve a simple equation \(z^2 + 6iz - 5z - 9i - 1 = 0\)

First of all, we need to group all coefficients and then apply traditional way for solving square equations via determinant.

\[ z^2 + 6iz - 5z - 9i - 1 = 0 \\ z^2 + (6i - 5)z - (9i + 1) = 0 \\ Z_{1,2} = \frac{-(6i - 5) \pm \sqrt{(6i - 5)^2 + 4(9i + 1)}}{2} \\ D = (6i - 5)^2 + 4(9i + 1) = \cdot = (6i - 5)^2 + 36i + 4 = -7 - 24i\]

I will not show here, how to find square roots from this determinant, but will refer you to a previous lecture about finding square roots from complex numbers. Yes, here we need to use this formula or derived it again. In the video, I will show you again how to derive this equation, but here I will give you the results.

\[\sqrt{-7 - 24i} = 3 - 4i\]

therefore

\[Z_{1,2} = \frac{-6i + 5 \pm (3 - 4i)}{2} \]

After simplification we will have:

\[ z^2 + 6iz - 5z - 9i - 1 = 0 \\ Z_1 = 4 - 5i; Z_2 = 1 - i\]

For more detailed explanation, please go to watch our video:


Published: 2023-05-09 23:48:26
Updated: 2023-05-09 23:59:19

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