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Chapter 11: Problem 66

Find the indicated complex roots. Express your answers in polar form and then convert them into rectangular form. the two square roots of \(z=-25 i\)

Short Answer

Expert verified
The square roots are \(-\frac{5\sqrt{2}}{2} + i\frac{5\sqrt{2}}{2}\) and \(\frac{5\sqrt{2}}{2} - i\frac{5\sqrt{2}}{2}\).

Step by step solution

01

Convert to Polar Form

First, express the complex number in polar form. The given number is \( z = -25i \). In polar form, a complex number is represented as \( z = r(\cos\theta + i\sin\theta) \). Here, \( r \) is the modulus and \( \theta \) is the argument of the complex number. The modulus \( r = \sqrt{(-25)^2} = 25 \). The argument \( \theta \) corresponds to the angle, and since the number is on the negative imaginary axis, \( \theta = \frac{3\pi}{2} \). Thus, \( z = 25(\cos\frac{3\pi}{2} + i\sin\frac{3\pi}{2}) \).
02

Find the Roots in Polar Form

To find the square roots, we use the formula for roots of complex numbers in polar form: \( w_k = r^{1/n} \left( \cos \left( \frac{\theta + 2k\pi}{n} \right) + i \sin \left( \frac{\theta + 2k\pi}{n} \right) \right) \). Here, \( n = 2 \) for square roots, \( r = 25 \), and \( \theta = \frac{3\pi}{2}\). Thus, \( r^{1/2} = \sqrt{25} = 5 \). Now, substitute and evaluate for \( k = 0 \) and \( k = 1 \).
03

Calculate for k = 0

Substitute \( k = 0 \) into the formula: \[ w_0 = 5 \left( \cos \left( \frac{3\pi}{4} \right) + i \sin \left( \frac{3\pi}{4} \right) \right) \]Evaluating the trigonometric values, we have \( \cos\frac{3\pi}{4} = -\frac{\sqrt{2}}{2} \) and \( \sin\frac{3\pi}{4} = \frac{\sqrt{2}}{2} \). Thus,\[ w_0 = 5 \left(-\frac{\sqrt{2}}{2} + i \frac{\sqrt{2}}{2}\right) = -\frac{5\sqrt{2}}{2} + i\frac{5\sqrt{2}}{2} \].
04

Calculate for k = 1

Substitute \( k = 1 \) into the formula: \[ w_1 = 5 \left( \cos \left( \frac{7\pi}{4} \right) + i \sin \left( \frac{7\pi}{4} \right) \right) \]Evaluating the trigonometric values, we have \( \cos\frac{7\pi}{4} = \frac{\sqrt{2}}{2} \) and \( \sin\frac{7\pi}{4} = -\frac{\sqrt{2}}{2} \). Thus,\[ w_1 = 5 \left(\frac{\sqrt{2}}{2} - i \frac{\sqrt{2}}{2}\right) = \frac{5\sqrt{2}}{2} - i\frac{5\sqrt{2}}{2} \].
05

Final Results in Rectangular Form

The two square roots of \( z = -25i \) in rectangular form are:1. \(-\frac{5\sqrt{2}}{2} + i\frac{5\sqrt{2}}{2} \)2. \(\frac{5\sqrt{2}}{2} - i\frac{5\sqrt{2}}{2} \)

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Key Concepts

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

Polar Form
Complex numbers can be expressed in either rectangular form or polar form. When dealing with the polar form, we represent a complex number as \( z = r(\cos\theta + i\sin\theta) \). Here, \( r \) is the modulus, which represents the distance from the origin to the point representing the complex number on the complex plane. \( \theta \) is the argument, indicating the direction or angle of the line connecting the point to the origin.
  • The modulus \( r \) is calculated using the formula: \( r = \sqrt{a^2 + b^2} \) for a complex number \( a + bi \).
  • The argument \( \theta \) is often found using trigonometric identities, based on the location of the complex number on the complex plane.
Representing numbers in polar form is particularly useful when performing multiplication, division, or finding powers and roots of complex numbers, as it simplifies these operations significantly.
Rectangular Form
The rectangular form of a complex number is simply its standard format: \( a + bi \), where \( a \) is the real part and \( bi \) is the imaginary part. This form is intuitive as it straightforwardly tells you both components of the number right away.
  • To convert from polar to rectangular form, evaluate the \( \cos \theta \) and \( \sin \theta \) components, then multiply by \( r \).
  • This conversion needs knowledge of trigonometric values, such as \( \cos \frac{3\pi}{4} = -\frac{\sqrt{2}}{2} \) and \( \sin \frac{3\pi}{4} = \frac{\sqrt{2}}{2} \).
Understanding both forms aids in applying different mathematical operations suited for specific applications, making it essential for studying complex numbers.
Square Roots
Finding square roots of complex numbers involves expressing them in polar form first. Once in polar form, we use the formula for extracting \( n \)-th roots: \[ w_k = r^{1/n} \left( \cos \left( \frac{\theta + 2k\pi}{n} \right) + i \sin \left( \frac{\theta + 2k\pi}{n} \right) \right) \].Here, \( n \) represents the root; for square roots, \( n = 2 \). This formula efficiently finds all roots by iterating through \( k \) values starting from 0.
  • Calculate \( r^{1/2} \), which simplifies to the square root of \( r \).
  • For each \( k \), compute the angle adjustments and find out the distinct roots.
Each calculated root can be converted back to rectangular form if needed, providing insight into the geometric interpretation on the complex plane.

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Most popular questions from this chapter

The table below lists the average temperature of Lake Erie as measured in Cleveland, Ohio on the first of the month for each month during the years \(1971-2000 .^{19}\) For example, \(t=3\) represents the average of the temperatures recorded for Lake Erie on every March 1 for the years 1971 through 2000 . $$ \begin{array}{|l|r|r|r|r|r|r|r|r|r|r|r|r|} \hline \text { Month } & & & & & & & & & & & & \\ \text { Number, } t & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 \\ \hline \begin{array}{l} \text { Temperature } \\ \left({ }^{\circ} \mathrm{F}\right), T \end{array} & 36 & 33 & 34 & 38 & 47 & 57 & 67 & 74 & 73 & 67 & 56 & 46 \\ \hline \end{array} $$ (a) Using the techniques discussed in Example 11.1.2, fit a sinusoid to these data. (b) Using a graphing utility, graph your model along with the data set to judge the reasonableness of the fit. (c) Use the model you found in part 8 a to predict the average temperature recorded for Lake Erie on April \(15^{\text {th }}\) and September \(15^{\text {th }}\) during the years \(1971-2000 .^{20}\) (d) Compare your results to those obtained using a graphing utility.

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In Exercises \(41-50\), use set-builder notation to describe the polar region. Assume that the region contains its bounding curves. The region in Quadrant I which lies inside both the circle \(r=3\) as well as the rose \(r=6 \sin (2 \theta)\).

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