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Chapter 3: Problem 14

Find the three cube roots of \(3-2 i\).

Short Answer

Expert verified
The cube roots of \(3-2 i\) are approximately \(1.87939-0.68465i\), \(-0.532433+1.8794i\) and \(-1.34696-1.19475i\).

Step by step solution

01

Convert to Polar Form

First, convert the complex number \(3-2 i\) to polar form. The magnitude \(r\) is calculated as \(r=\sqrt{(3)^2 + (-2)^2} = \sqrt{13}\). The argument \(\theta\), in radian measures, satisfies \(\cos(\theta)=3/\sqrt{13}\) and \(\sin(\theta)=-2/\sqrt{13}\). \(\theta\) is thus \(arccos(3/\sqrt{13})\) which ends up in the fourth quadrant.
02

Apply De Moivre's Theorem

Next, apply De Moivre's theorem, which states that if \(z = r (cos \theta + i sin \theta)\), then its \(n\)-th roots are \(z_k = r^{1/n} (cos ((\theta+2k\pi)/n) + i sin ((\theta+2k\pi)/n))\), where \(k=0, 1, 2,... n-1\). Since we are finding the cube roots \(n=3\), so \(k=0,1,2\). Plug in \(r^{1/n}\), \(n\) and \(\theta\) with these \(k\) values.
03

Convert Back to Cartesian Form

Finally, calculate the roots and then convert the roots back to Cartesian form. It might be useful to use engine to compute the cosine and sine values.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Cube Roots
When we talk about cube roots, we refer to the operation that finds a number which, when multiplied by itself three times, gives the original number. In the context of complex numbers, finding cube roots can sometimes be tricky due to their imaginary component.

However, using the polar form and De Moivre's theorem vastly simplifies this process. You start by transforming your complex number into its polar form. The polar form gives a clear picture of the number's magnitude and its angle, or argument, in the complex plane.

Once you have the polar form, you can systematically derive the cube roots, making them easy to find and understand.
Polar Form
The polar form of a complex number is a way of expressing a complex number in terms of its magnitude and angle in the complex plane. It helps visualize and calculate operations with complex numbers.

To convert a complex number into polar form, you need to find both its magnitude and argument.
  • Magnitude: This is calculated using the formula \( r = \sqrt{a^2 + b^2} \), where \( a \) and \( b \) are the real and imaginary parts of the number, respectively.

  • Argument: This is the angle \( \theta \) the complex number makes with the positive x-axis, calculated using trigonometric functions. For the number \( 3 - 2i \), it involves \( \arccos(3/\sqrt{13}) \).

This form is especially useful because operations like finding powers or roots of complex numbers become more straightforward with it.
De Moivre's Theorem
De Moivre's Theorem is a powerful mathematical tool that simplifies exponentiation of complex numbers. It states that for a complex number in polar form, \( z = r(\cos \theta + i \sin \theta) \), its powers and roots can be easily calculated by manipulating its magnitude and angle. Here鈥檚 how it works:

  • To find the nth roots, use the formula \( z_k = r^{1/n} \left( \cos \left( \frac{\theta + 2k\pi}{n} \right) + i\sin \left( \frac{\theta + 2k\pi}{n} \right) \right) \), where \( n \) is the root you're finding, and \( k = 0, 1, ..., n-1 \).

  • This approach decomposes the task of finding roots into handling simpler trigonometric calculations and makes the solution neat.

By employing De Moivre's theorem, we can find all nth roots of a complex number with precision and ease, as shown in how we derived the cube roots of \( 3-2i \). It minimizes errors and simplifies the complexities involved in complex arithmetic.

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Most popular questions from this chapter

The solutions to the Schr枚dinger equation for the electron in a hydrogen atom have three quantum numbers associated with them, called \(n, l\), and \(m\), and these solutions are denoted by \(\psi_{n l m}\). (a) The \(\psi_{210}\) function is given by $$ \psi_{210}=\frac{1}{4 \sqrt{2 \pi}}\left(\frac{1}{a_{0}}\right)^{3 / 2} \frac{r}{a_{0}} e^{-r / 2 a_{0}} \cos (\theta) $$ where \(a_{0}=0.529 \times 10^{-10} \mathrm{~m}\) is called the Bohr radius. Write this function in terms of Cartesian coordinates. (b) The \(\psi_{211}\) function is given by $$ \psi_{211}=\frac{1}{8 \sqrt{\pi}}\left(\frac{1}{a_{0}}\right)^{3 / 2} \frac{r}{a_{0}} e^{-r / 2 a_{0}} \sin (\theta) e^{i \phi} $$ Write an expression for the magnitude of this complex function. (c) The \(\psi_{211}\) function is sometimes called \(\psi_{2 p 1}\). Write expressions for the real and imaginary parts of the function, which are proportional to the related functions. Which are called \(\psi_{2 p x}\) and \(\psi_{2 p y}\).

Express the equation \(y=b\), where \(b\) is a constant, in plane polar coordinates.

Obtain the famous formulas $$ \begin{aligned} &\sin (\phi)=\frac{e^{i \phi}-e^{-i \phi}}{2 i}=I\left(e^{i \phi}\right), \\ &\cos (\phi)=\frac{e^{i \phi}+e^{-i \phi}}{2}=R\left(e^{i \phi}\right) . \end{aligned} $$

A surface is represented in cylindrical polar coordinates by the equation \(z=\rho^{2}\). Describe the shape of the surface.

{ If } z=\left(\frac{3+2 i}{4+5 i}\right)^{2}, \text { find } R(z), I(z), r, \text { and } \phi
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