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The equation $x^2 + 2x = i - 2x$ has two complex solutions. Determine the product of their real parts.

Aug 23, 2023

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I know very little about the complex numbers realm of mathematics, but I suppose I know enough to find the answer to this particular problem although I doubt my method is the cleanest. or even the simplest because typically problems involving complex numbers can have intricate solutions. Regardless, I will present my thought process, and I hope you can follow the logic. I first started by completing the square of this complex quadratic, which gave me some insight about the real part of the product of the roots.

$$x^2 + 2x = i - 2x \\ x^2 + 4x = i \\ x^2 + 4x + 4 = i + 4 \\ (x + 2)^2 = i + 4 \\ |x + 2| = \sqrt{i + 4} \\ x_1 = -2 + \sqrt{i + 4} \text{ or } x_2 = -2 - \sqrt{i + 4}$$

Unfortunately, we are left with an undesirable term, $$\sqrt{i + 4}$$. We will have to extract the real part from this expression to be able to find the product of the real parts of the complex solutions. For now, I will just assume I know what the real part is and simplify as much as possible while including the $$-2$$ term since $$-2 \in \mathbb{R}$$.

$$\Re(x_1x_2) = \left(-2 + \Re\left(\sqrt{i + 4}\right)\right)\left(-2 - \Re\left(\sqrt{i + 4}\right)\right) \\ \Re(x_1x_2) = 4 - \left(\Re\left(\sqrt{i + 4}\right)\right)^2$$

Once again, we have to extract the real part out of $$\sqrt{i + 4}$$ somehow. The complex numbers are algebraically closed in almost all operations except arguably division because of issues arising from division by zero. As a result, we know that the square root of a complex number will yield another complex number. Because of the algebraic closure, we can write $$\sqrt{i + 4} = a + bi$$ where $$a$$ is the real part and $$b$$ is the imaginary part and solve for each of its components.

$$\sqrt{i + 4} = a + bi \\ i + 4 = a^2 + 2abi + b^2i^2 \\ i + 4 = 2abi + (a^2 - b^2)$$

Now, we can write a system of equations and solve enough to extract the desired variable. In this case, we only care about $$a$$ since that represents the real part of $$\sqrt{i + 4}$$

$$\begin{cases} a^2 - b^2 = 4 \\ 2ab = 1 \end{cases} \\ \begin{cases} b^2 = a^2 - 4 \\ 4a^2b^2 = 1 \end{cases}$$

Now, we have an easy substitution for $$a^2$$, which will allow us to solve for $$a$$ and finish this problem.

$$4a^2(a^2 - 4) = 1 \\ 4(a^4 - 4a^2) = 1$$

This is a quartic polynomial with only even-numbered exponents on the a-variable. This means we can treat this equation like a quadratic by letting $$y = a^2$$ and then solving it again by another version of completing the square.

$$4(y^2 - 4y) = 1 \\ 4(y^2 - 4y + 4) = 1 + 4*4 \\ 4(y - 2)^2 = 17 \\ (y - 2)^2 = \frac{17}{4} \\ y - 2 = \pm \frac{\sqrt{17}}{2} \\ a_1^2 = 2 + \frac{\sqrt{17}}{2} \text{ or } a_2^2 = 2 - \frac{\sqrt{17}}{2}$$

There is no need to go any further than this since we only need the square of the real part. However, we now need to choose between these two viable solutions. To choose, first recall that the a-variable represents $$\Re\left(\sqrt{i + 4}\right)$$, so $$a \in \mathbb{R}$$. Since $$2 + \frac{\sqrt{17}}{2} > 0, a_1 \in \mathbb{R}$$. However, since $$2 - \frac{\sqrt{17}}{2} < 0, a_2 \not\in \mathbb{R}$$. This means that $$a^2 = 2 + \frac{\sqrt{17}}{2}$$ is the only correct solution for this particular situation. We finally have the information we seeked from the beginning of the problem.

\begin{align*} \Re(x_1x_2) &= 4 - \left(\Re\left(\sqrt{i + 4}\right)\right)^2 \\ &= 4 - \left(2 + \frac{\sqrt{17}}{2} \right) \\ &= 2 - \frac{\sqrt{17}}{2} \end{align*}

We are done!

Aug 25, 2023
edited by The3Mathketeers  Aug 25, 2023