Solving Complex Cubic Equations: A Step-by-Step Guide

Hey guys! Let's dive into the fascinating world of complex numbers and cubic equations. We'll be tackling a specific problem today: considering the equation (E): $z^3 = 2 + 11i$ in the complex plane C. This means we're looking for complex numbers $z$ that, when cubed, give us the result $2 + 11i$. Don't worry, it sounds a bit intimidating, but we'll break it down step-by-step to make it super clear and easy to follow. We'll also explore how to actually solve this kind of equation and what it all means.

Verifying a Solution

First up, we need to verify whether a given complex number is a solution to our equation. This is a common task, so understanding this process is crucial. The problem provides us with a potential solution: $z_0 = 2 + i$. Our job? To check if this value satisfies the equation (E). This means we'll substitute $z_0$ into the equation and see if both sides match up. Let's roll up our sleeves and do some calculations. We will substitute $z_0$ in $(E): z^3 = 2+11i$. Doing the substitution $z_0^3 = (2+i)^3$, which is $(2+i)^2(2+i)$. We first expand $(2+i)^2$ which gives us $4 + 4i + i^2$. Since $i^2 = -1$, the expansion is equal to $4 + 4i - 1 = 3 + 4i$. Next, we multiply it by $(2+i)$. This gives us $(3+4i)(2+i)$ which is $6 + 3i + 8i + 4i^2$. Again since $i^2 = -1$, the result is $6+11i - 4$, finally, the result is $2 + 11i$. This is the same as the right side of the original equation. Therefore, we can conclude that $z_0 = 2+i$ is indeed a solution to the equation (E). This means we've found at least one complex number that works with our equation. This first step might seem simple, but it really showcases the basics of working with complex numbers. You see how we deal with the powers of $i$ and how we simplify the expression? This process is fundamental to more complex calculations down the line. Remember that the real and imaginary parts are separate entities, and must be handled according to the rules of complex arithmetic. Also, this shows us that, complex number solutions can be verified by direct substitution and simplification. This ensures we understand the basic rules and mechanics of complex numbers, before we move on.

Solving the Cubic Equation in C

Alright, now for the main course: solving the cubic equation $(E): z^3 = 2 + 11i$ in the complex plane C. The previous part gave us one solution $z_0 = 2 + i$. Now we need to figure out if there are any other solutions and how to find them. Because it's a cubic equation, we know (by the Fundamental Theorem of Algebra) that it should have three solutions (counting multiplicities). Since we already know one, let's figure out the other two. There are several approaches to solving cubic equations. Since we already know one solution, we can factorize it by dividing the cubic equation by the expression with $z_0$. But that's quite challenging. Let's explore a more accessible method. We can rewrite the equation as $z^3 - (2 + 11i) = 0$. The first step is to use the already known solution $z_0$. Since $z_0$ is a solution, we know that $(z - z_0)$ is a factor of the equation. Now, we can do a polynomial division or using the fact that if we know one root, the other roots can be found by other means. One way to solve a cubic equation is to use the polar form. If $z$ is a solution to $z^3 = 2 + 11i$, then we write $z$ as $z = r(\cos(\theta) + i\sin(\theta))$, where $r$ is the magnitude of $z$ and $\theta$ is the argument (angle) of $z$. Then $z^3 = r^3(\cos(3\theta) + i\sin(3\theta))$. Also we write $2 + 11i$ in polar form. First we have the magnitude, which is $\sqrt{2^2 + 11^2} = \sqrt{125} = 5\sqrt{5}$. We write the angle as $\phi$, then $\cos(\phi) = \frac{2}{5\sqrt{5}}$ and $\sin(\phi) = \frac{11}{5\sqrt{5}}$. Therefore, we have $2 + 11i = 5\sqrt{5}(\cos(\phi) + i\sin(\phi))$. By equating both the equations of $z^3$, we have $r^3 = 5\sqrt{5}$, which leads to $r = \sqrt{5}$. The other equation we get is $3\theta = \phi + 2k\pi$, with $k$ being an integer, this will lead to $\theta = \frac{\phi}{3} + \frac{2k\pi}{3}$. Now, we know that one solution is $2+i$, which, if you do the math in polar form, is equivalent. If we calculate for $k = 1$, we have $\theta = \frac{\phi}{3} + \frac{2\pi}{3}$ and $k=2$, we have $\theta = \frac{\phi}{3} + \frac{4\pi}{3}$. So that means we have three solutions in the polar form. Another approach would be to convert it to the rectangular form, but that will be even more challenging. In summary, solving cubic equations involving complex numbers requires careful application of algebraic techniques, knowledge of complex number properties, and a clear understanding of the goal. By systematically breaking down the problem, we can arrive at the solutions, even if they involve a bit of calculation and critical thinking. Remember that practice makes perfect. So, try solving similar problems to get better at working with complex numbers.

Finding all roots

We have already found one root which is $z_0 = 2 + i$. So, now we can express the equation as $z^3 - (2 + 11i) = 0$, we can express it as $(z - (2+i))(z^2 + a z + b) = 0$. If we do the calculation $z^2 + a z + b = \frac{z^3 - (2 + 11i)}{z - (2+i)}$. By expanding $(z - (2+i))(z^2 + a z + b)$, we have $z^3 + az^2 + bz - (2+i)z^2 - (2+i)az - (2+i)b$. Then, we can see that $a = -(2+i)$ and $b = -5$. We can verify the results by expanding again $(z - (2+i))(z^2 - (2+i) z -5) = z^3 - (2+i)z^2 - 5z - (2+i)z^2 + (2+i)^2z + 5(2+i) = z^3 - (4+2i)z^2 + (-5 + 4+3i)z + (10+5i) = z^3 - (4+2i)z^2 + (-1+3i)z + 10+5i$. The polynomial division gives us $z^2 - (2+i)z -5 = 0$. By using the quadratic formula $z = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{2+i \pm \sqrt{(2+i)^2 + 20}}{2} = \frac{2+i \pm \sqrt{3+4i+20}}{2}$. So, this is equivalent to $\frac{2+i \pm \sqrt{23+4i}}{2}$. We can rewrite $23+4i$ as $x^2 - y^2 = 23$ and $2xy = 4$, so $xy = 2$. So we can have $x = 2$ and $y = 1$, this gives us $x^2 + y^2 = 5$. So we can rewrite $23+4i$ as $(x+iy)^2$, with $x^2 - y^2 = 23$ and $2xy = 4$, so $(x+iy)^2 = 23+4i$. So $x+iy = \sqrt{23+4i}$. Solving those two equations, we get $x = 2\sqrt{6}$ and $y = \frac{1}{\sqrt{6}}$. Which leads to a solution equal to $2+i \pm \sqrt{23+4i}$.

So, the solution can also be written as:

• $z_1 = 2+i$
• $z_2 = \frac{2+i + \sqrt{23+4i}}{2}$
• $z_3 = \frac{2+i - \sqrt{23+4i}}{2}$

This represents the three roots of the cubic equation.

Conclusion

So there you have it! We've successfully verified a solution, explored different strategies, and solved the cubic equation $(E): z^3 = 2 + 11i$ in the complex plane. We also demonstrated that the cubic equation can be solved by the complex roots by converting it into polar and rectangular forms. This journey through complex numbers and cubic equations hopefully has given you a deeper appreciation for the beauty and elegance of mathematics. Keep practicing, keep exploring, and never stop asking "why?" The world of math is full of amazing discoveries. Keep up the great work, and happy solving!