/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 48 Use \(z=-\frac{3 \sqrt{3}}{2}+\f... [FREE SOLUTION] | 91影视

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Chapter 11: Problem 48

Use \(z=-\frac{3 \sqrt{3}}{2}+\frac{3}{2} i\) and \(w=3 \sqrt{2}-3 i \sqrt{2}\) to compute the quantity. Express your answers in polar form using the principal argument. $$ \frac{z^{2}}{w} $$

Short Answer

Expert verified
\( \frac{3}{8} \text{cis} \left(-\frac{\pi}{12}\right) \)

Step by step solution

01

Calculate the square of z

Given the complex number \( z = -\frac{3 \sqrt{3}}{2} + \frac{3}{2} i \), we need to square it: \( z^2 = \left(-\frac{3 \sqrt{3}}{2} + \frac{3}{2} i\right)^2 \). Use the formula for squaring a complex number in the form of \( (a + bi)^2 = a^2 - b^2 + 2abi \). Here, \( a = -\frac{3 \sqrt{3}}{2} \) and \( b = \frac{3}{2} \). Substitute these values to get:\[z^2 = \left(-\frac{3\sqrt{3}}{2}\right)^2 - \left(\frac{3}{2}\right)^2 + 2 \left(-\frac{3\sqrt{3}}{2}\right) \left(\frac{3}{2}\right) i\]Calculate:\[= \frac{27}{4} - \frac{9}{4} - 9\sqrt{3} i\]Which simplifies to:\[z^2 = \frac{18}{4} - 9\sqrt{3} i = \frac{9}{2} - 9\sqrt{3} i\]
02

Express w in polar form

The given complex number\( w = 3 \sqrt{2} - 3i \sqrt{2} \) can be written in polar form. First, find the modulus \( r \) using:\[r = \sqrt{(3\sqrt{2})^2 + (-3\sqrt{2})^2} \]Calculate:\[= \sqrt{18 + 18} = \sqrt{36} = 6\]Now, find the argument \( \theta \) using:\[\theta = \tan^{-1}\left(\frac{-3\sqrt{2}}{3\sqrt{2}}\right) = \tan^{-1}(-1)\]Since \( w \) lies in the fourth quadrant, \( \theta = -\frac{\pi}{4} \). Therefore, \( w \) in polar form is:\[w = 6 \text{cis} \left(-\frac{\pi}{4}\right)\]
03

Compute \(\frac{z^2}{w}\)

Now that we have \( z^2 = \frac{9}{2} - 9\sqrt{3} i \) and \( w = 6 \text{cis} (-\frac{\pi}{4}) \), calculate the division:\[\frac{z^2}{w} = \frac{\left(\frac{9}{2} - 9\sqrt{3} i\right)}{6 \text{cis} \left(-\frac{\pi}{4}\right)}\]First, simplify the fraction:\[= \frac{\frac{9}{2} - 9\sqrt{3} i}{6} = \frac{3}{4} - \frac{3\sqrt{3}}{2} i\]Find the modulus of the numerator:\[= \sqrt{\left(\frac{3}{4}\right)^2 + \left(\frac{3\sqrt{3}}{2}\right)^2} = \sqrt{\frac{9}{16} + \frac{27}{4}} = \sqrt{\frac{117}{16}} = \frac{9}{4}\]The argument (\( \phi\)) of the numerator can be found using:\[\tan(\phi) = \frac{-\frac{3\sqrt{3}}{2}}{\frac{3}{4}} = -\sqrt{3}\]\( \phi = -\frac{\pi}{3} \) (third quadrant). Therefore the polar form before division is:\[\frac{9}{4}\text{cis}(-\frac{\pi}{3})\]Now divide:\[\frac{\frac{9}{4}}{6} \text{cis} \left(-\frac{\pi}{3} - \left(-\frac{\pi}{4}\right)\right) = \frac{3}{8} \text{cis} \left(-\frac{\pi}{3} + \frac{\pi}{4}\right)\]Calculate the new argument:\[= \frac{3}{8} \text{cis} \left(-\frac{\pi}{12}\right)\]
04

Final Polar Form

The final expression for \( \frac{z^2}{w} \) in polar form is:\[\frac{3}{8} \text{cis} \left(-\frac{\pi}{12}\right)\]

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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 not only in rectangular form but also in polar form. This form is extremely useful when multiplying, dividing, or finding powers and roots of complex numbers. In polar form, a complex number is represented using a magnitude (modulus) and an angle (argument). This is given by:\[z = r \text{cis} \theta\]where \(r\) is the modulus of the complex number and \(\theta\) is the argument. The term "cis" stands for cosine plus i times sine, succinctly expressing the exponential form: \[z = r(\cos \theta + i \sin \theta)\]Working with complex numbers in polar form simplifies certain operations, such as multiplication and division, as the exponents can be directly manipulated.
Complex Division
When dividing complex numbers, especially in their polar form, the process becomes simpler than dealing with rectangular form. Consider two complex numbers, \(z_1 = r_1 \text{cis} \theta_1\) and \(z_2 = r_2 \text{cis} \theta_2\). To divide \(z_1\) by \(z_2\), we perform:
  • Divide their moduli: \(\frac{r_1}{r_2}\)
  • Subtract their arguments: \(\theta_1 - \theta_2\)
Thus, the result of division in polar form will be:\[\frac{z_1}{z_2} = \left(\frac{r_1}{r_2}\right) \text{cis} (\theta_1 - \theta_2)\]This method highlights the beauty of polar form, as it transforms division into simple arithmetic operations on the modulus and argument of the complex numbers involved.
Complex Modulus
The modulus of a complex number is the distance from the origin to the point in the complex plane. It's calculated using the formula:\[|z| = \sqrt{a^2 + b^2}\]where \(z = a + bi\). For example, the modulus of our given complex number \( w = 3\sqrt{2} - 3i\sqrt{2} \) involves calculating:
  • Squaring both the real and imaginary parts: \((3\sqrt{2})^2 + (-3\sqrt{2})^2\)
  • Summing these squares: \(18 + 18 = 36\)
  • Taking the square root: \(\sqrt{36} = 6\)
The modulus gives rise to the term \(r\) in polar representation, crucial for plotting and performing operations in the polar coordinate system.
Argument of a Complex Number
The argument of a complex number gives the angle formed with the positive real axis in the complex plane. Calculating this angle can be crucial when expressing a complex number in polar form or performing trigonometric operations. Given a complex number \(z = a + bi\), the argument \(\theta\) is found using:\[\theta = \tan^{-1}\left(\frac{b}{a}\right)\]This is often adjusted depending on the quadrant in which the complex number lies. For example, the complex number \(w = 3\sqrt{2} - 3i\sqrt{2}\) lies in the fourth quadrant of the complex plane, changing our base argument to correspond to:
  • \(\tan^{-1}(-1) = -\frac{\pi}{4}\), considering its position.
Understanding how to correctly determine this angle is important for accurate representation and manipulation of complex numbers.

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Most popular questions from this chapter

Convert the equation from polar coordinates into rectangular coordinates. $$ r=1+\sin (\theta) $$

Suppose Taylor fills her wagon with rocks and must exert a force of 13 pounds to pull her wagon across the yard. If she maintains a \(15^{\circ}\) angle between the handle of the wagon and the horizontal, compute how much work Taylor does pulling her wagon 25 feet. Round your answer to two decimal places.

For the given vector \(\vec{v}\), find the magnitude \(\|\vec{v}\|\) and an angle \(\theta\) with \(0 \leq \theta<360^{\circ}\) so that \(\vec{v}=\|\vec{v}\|\langle\cos (\theta), \sin (\theta)\rangle\) (See Definition 11.8.) Round approximations to two decimal places. $$ \vec{v}=\hat{\imath}+\hat{\jmath} $$

In Exercises \(25-39\), find a parametric description for the given oriented curve. the directed line segment from \((-2,-1)\) to \((3,-4)\)

We know that \(|x+y| \leq|x|+|y|\) for all real numbers \(x\) and \(y\) by the Triangle Inequality established in Exercise 36 in Section 2.2. We can now establish a Triangle Inequality for vectors. In this exercise, we prove that \(\|\vec{u}+\vec{v}\| \leq\|\vec{u}\|+\|\vec{v}\|\) for all pairs of vectors \(\vec{u}\) and \(\vec{v}\). (a) (Step 1) Show that \(\|\vec{u}+\vec{v}\|^{2}=\|\vec{u}\|^{2}+2 \vec{u} \cdot \vec{v}+\|\vec{v}\|^{2}\). 6 (b) (Step 2) Show that \(|\vec{u} \cdot \vec{v}| \leq\|\vec{u}\|\|\vec{v}\| .\) This is the celebrated Cauchy-Schwarz Inequality. (Hint: To show this inequality, start with the fact that \(|\vec{u} \cdot \vec{v}|=|\|\vec{u}\|\|\vec{v}\| \cos (\theta)|\) and use the fact that \(|\cos (\theta)| \leq 1\) for all \(\theta\).) (c) (Step 3) Show that \(\|\vec{u}+\vec{v}\|^{2}=\|\vec{u}\|^{2}+2 \vec{u} \cdot \vec{v}+\|\vec{v}\|^{2} \leq\|\vec{u}\|^{2}+2|\vec{u} \cdot \vec{v}|+\|\vec{v}\|^{2} \leq\|\vec{u}\|^{2}+\) \(2\|\vec{u}\|\|\vec{v}\|+\|\vec{v}\|^{2}=(\|\vec{u}\|+\|\vec{v}\|)^{2}\) (d) (Step 4) Use Step 3 to show that \(\|\vec{u}+\vec{v}\| \leq\|\vec{u}\|+\|\vec{v}\|\) for all pairs of vectors \(\vec{u}\) and \(\vec{v}\). (e) As an added bonus, we can now show that the Triangle Inequality \(|z+w| \leq|z|+|w|\) holds for all complex numbers \(z\) and \(w\) as well. Identify the complex number \(z=a+b i\) with the vector \(u=\langle a, b\rangle\) and identify the complex number \(w=c+d i\) with the vector \(v=\langle c, d\rangle\) and just follow your nose!
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