Find All Solutions In \mathbb{C} Of The Equation Z^4 =1-2i

$$$$

Introduction

In this article, we will delve into the world of abstract algebra and explore the solutions to the equation $z^4 =1-2i$ in the complex numbers $\mathbb{C}$. The equation $z^4 =1-2i$ is a quartic equation, and solving it will involve some advanced techniques from complex analysis and algebra.

Background

To begin, let's recall some basic facts about complex numbers. A complex number $z$ can be written in the form $z = a + bi$, where $a$ and $b$ are real numbers and $i$ is the imaginary unit, which satisfies $i^2 = -1$. The set of all complex numbers is denoted by $\mathbb{C}$.

Converting the equation to polar form

To solve the equation $z^4 =1-2i$, we can start by converting it to polar form. Let $z = re^{i\theta}$, where $r$ is the magnitude of $z$ and $\theta$ is the argument of $z$. Then, we have:

$$z^4 = (re^{i\theta})^4 = r^4e^{i4\theta}$$

We also have:

$$1-2i = \sqrt{5}e^{-i\arctan(2)}$$

Equating the two expressions, we get:

$$r^4e^{i4\theta} = \sqrt{5}e^{-i\arctan(2)}$$

Finding the magnitude and argument

From the equation above, we can see that the magnitude of $z^4$ is equal to the magnitude of $1-2i$, which is $\sqrt{5}$. Therefore, we have:

$$r^4 = \sqrt{5}$$

Taking the fourth root of both sides, we get:

$$r = \sqrt[4]{\sqrt{5}}$$

Now, let's consider the argument of $z^4$. We have:

$$4\theta = -\arctan(2) + 2k\pi$$

where $k$ is an integer. Dividing both sides by 4, we get:

$$\theta = -\frac{\arctan(2)}{4} + \frac{k\pi}{2}$$

Finding the solutions

Now that we have the magnitude and argument of $z^4$, we can find the solutions to the equation $z^4 =1-2i$. We have:

$$z = re^{i\theta} = \sqrt[4]{\sqrt{5}}e^{i\left(-\frac{\arctan(2)}{4} + \frac{k\pi}{2}\right)}$$

where $k$ is an integer.

Listing the solutions

To find the solutions, we can plug in different values of $k$ into the expression above. We get:

$$z_1 = \sqrt[4]{\sqrt{5}}e^{i\left(-\frac{\arctan(2)}{4} + \frac{\pi}{2}\right)}$$

$$z_2 = \sqrt[4]{\sqrt{5}}e^{i\left(-\frac{\arctan(2)}{4} + \frac{3\pi}{2}\right)}$$

$$z_3 = \sqrt[4]{\sqrt{5}}e^{i\left(-\frac{\arctan(2)}{4} + \pi\right)}$$

$$z_4 = \sqrt[4]{\sqrt{5}}e^{i\left(-\frac{\arctan(2)}{4} + \frac{5\pi}{2}\right)}$$

Simplifying the solutions

We can simplify the solutions by using the fact that $e^{i\pi} = -1$ and $e^{i\frac{\pi}{2}} = i$. We get:

$$z_1 = \sqrt[4]{\sqrt{5}}\left(-\cos\left(\frac{\arctan(2)}{4}\right) + i\sin\left(\frac{\arctan(2)}{4}\right)\right)$$

$$z_2 = \sqrt[4]{\sqrt{5}}\left(-\cos\left(\frac{\arctan(2)}{4}\right) - i\sin\left(\frac{\arctan(2)}{4}\right)\right)$$

$$z_3 = \sqrt[4]{\sqrt{5}}\left(\cos\left(\frac{\arctan(2)}{4}\right) + i\sin\left(\frac{\arctan(2)}{4}\right)\right)$$

$$z_4 = \sqrt[4]{\sqrt{5}}\left(\cos\left(\frac{\arctan(2)}{4}\right) - i\sin\left(\frac{\arctan(2)}{4}\right)\right)$$

Conclusion

In this article, we have found all the solutions to the equation $z^4 =1-2i$ in the complex numbers $\mathbb{C}$. The solutions are given by:

$$z_k = \sqrt[4]{\sqrt{5}}\left(\cos\left(\frac{\arctan(2)}{4} + \frac{k\pi}{2}\right) + i\sin\left(\frac{\arctan(2)}{4} + \frac{k\pi}{2}\right)\right)$$

where $k$ is an integer.

References

• [1] Ahlfors, L. V. (1979). Complex Analysis. McGraw-Hill.
• [2] Rudin, W. (1987). Real and Complex Analysis. McGraw-Hill.
• [3] Lang, S. (1999). Algebra. Springer-Verlag.

Further Reading

• [1] Complex Analysis by Lars V. Ahlfors
• [2] Real and Complex Analysis by Walter Rudin
• [3] Algebra by Serge Lang
  

$$$$

Introduction

In our previous article, we explored the solutions to the equation $z^4 =1-2i$ in the complex numbers $\mathbb{C}$. In this article, we will answer some frequently asked questions about the solutions to this equation.

Q: What is the magnitude of the solutions?

A: The magnitude of the solutions is given by $r = \sqrt[4]{\sqrt{5}}$.

Q: What is the argument of the solutions?

A: The argument of the solutions is given by $\theta = -\frac{\arctan(2)}{4} + \frac{k\pi}{2}$, where $k$ is an integer.

Q: How many solutions are there?

A: There are four solutions to the equation $z^4 =1-2i$.

Q: Can you list the solutions?

A: Yes, the solutions are given by:

$$z_k = \sqrt[4]{\sqrt{5}}\left(\cos\left(\frac{\arctan(2)}{4} + \frac{k\pi}{2}\right) + i\sin\left(\frac{\arctan(2)}{4} + \frac{k\pi}{2}\right)\right)$$

where $k$ is an integer.

Q: How do I simplify the solutions?

A: You can simplify the solutions by using the fact that $e^{i\pi} = -1$ and $e^{i\frac{\pi}{2}} = i$. This will give you the solutions in the form:

$$z_k = \sqrt[4]{\sqrt{5}}\left(-\cos\left(\frac{\arctan(2)}{4}\right) + i\sin\left(\frac{\arctan(2)}{4}\right)\right)$$

$$z_k = \sqrt[4]{\sqrt{5}}\left(-\cos\left(\frac{\arctan(2)}{4}\right) - i\sin\left(\frac{\arctan(2)}{4}\right)\right)$$

$$z_k = \sqrt[4]{\sqrt{5}}\left(\cos\left(\frac{\arctan(2)}{4}\right) + i\sin\left(\frac{\arctan(2)}{4}\right)\right)$$

$$z_k = \sqrt[4]{\sqrt{5}}\left(\cos\left(\frac{\arctan(2)}{4}\right) - i\sin\left(\frac{\arctan(2)}{4}\right)\right)$$

Q: What is the significance of the solutions?

A: The solutions to the equation $z^4 =1-2i$ are important in complex analysis and algebra. They can be used to study the properties of complex numbers and to solve other equations.

Q: How do I apply the solutions to real-world problems?

A: The solutions to the equation $z^4 =1-2i$ can be applied to real-world problems in fields such as engineering, physics, and computer science. For example, they can be used to model the behavior of electrical circuits, to study the properties of materials, and to develop new algorithms.

Conclusion

In this article, we have answered some frequently asked questions about the solutions to the equation $z^4 =1-2i$. We hope that this article has been helpful in understanding the solutions to this equation and in applying them to real-world problems.

References

• [1] Ahlfors, L. V. (1979). Complex Analysis. McGraw-Hill.
• [2] Rudin, W. (1987). Real and Complex Analysis. McGraw-Hill.
• [3] Lang, S. (1999). Algebra. Springer-Verlag.

Further Reading

• [1] Complex Analysis by Lars V. Ahlfors
• [2] Real and Complex Analysis by Walter Rudin
• [3] Algebra by Serge Lang