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Find the rectangular form of the given complex number. Use whatever identities are necessary to find the exact values. $$ z=\sqrt{10} \operatorname{cis}\left(\arctan \left(\frac{1}{3}\right)\right) $$

Short Answer

Expert verified
The rectangular form is \( z = 3 + i \).

Step by step solution

01

Understanding the Given Form

The complex number is given in polar form as \( z = r \operatorname{cis}(\theta) \) where \( r = \sqrt{10} \) and \( \theta = \arctan \left( \frac{1}{3} \right) \). Our goal is to convert this into rectangular form, \( z = x + yi \), where \( x = r\cos(\theta) \) and \( y = r\sin(\theta) \).
02

Calculating \( \cos(\theta) \) and \( \sin(\theta) \)

Since \( \theta = \arctan \left( \frac{1}{3} \right) \), we visualize a right triangle where the opposite side is 1 and the adjacent side is 3. Using the Pythagorean theorem, the hypotenuse is \( \sqrt{1^2 + 3^2} = \sqrt{10} \). Thus, \( \cos(\theta) = \frac{3}{\sqrt{10}} \) and \( \sin(\theta) = \frac{1}{\sqrt{10}} \).
03

Apply Values to Conversion Formulas

Now that we have \( \cos(\theta) = \frac{3}{\sqrt{10}} \) and \( \sin(\theta) = \frac{1}{\sqrt{10}} \), we can find \( x \) and \( y \) using the formulas \( x = r\cos(\theta) \) and \( y = r\sin(\theta) \). Thus, \( x = \sqrt{10} \times \frac{3}{\sqrt{10}} = 3 \) and \( y = \sqrt{10} \times \frac{1}{\sqrt{10}} = 1 \).
04

Combine to a Rectangular Form

Now combine \( x \) and \( y \) to write the complex number in rectangular form: \( z = x + yi = 3 + 1i \).

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Key Concepts

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

Polar to Rectangular Form
When dealing with complex numbers, it's often useful to convert between polar and rectangular forms. In polar form, a complex number is expressed as \( z = r \operatorname{cis}(\theta) \), where \( r \) is the modulus and \( \theta \) is the argument of the complex number. The term \( \operatorname{cis}(\theta) \) stands for \( \cos(\theta) + i \sin(\theta) \). This makes it easier to handle calculations involving magnitudes and angles. To convert from polar to rectangular form, we use the formulas \( x = r\cos(\theta) \) and \( y = r\sin(\theta) \). Here, \( x \) and \( y \) are the real and imaginary parts, respectively, of the rectangular representation \( z = x + yi \). This conversion is particularly useful in simplifying calculations and visualizing complex numbers on the complex plane. By breaking down the polar coordinates into their rectangular components, students gain a clearer understanding of complex numbers in a plane geometric context.
Trigonometric Identities
Trigonometric identities play a crucial role in converting complex numbers from polar to rectangular form. They allow us to calculate the precise values of \( \sin(\theta) \) and \( \cos(\theta) \), which are essential for conversion. For instance, the basic identities for sine and cosine in terms of a right triangle involve the opposite, adjacent sides, and the hypotenuse.
  • \( \cos(\theta) = \frac{\text{adjacent}}{\text{hypotenuse}} \)
  • \( \sin(\theta) = \frac{\text{opposite}}{\text{hypotenuse}} \)
In the example provided, \( \theta = \arctan \left( \frac{1}{3} \right) \), which makes \( \theta \) the angle in a triangle where the opposite side is 1 and the adjacent side is 3. By using the Pythagorean theorem, we determine the hypotenuse as \( \sqrt{10} \). Thus, \( \cos(\theta) \) and \( \sin(\theta) \) can be calculated as \( \frac{3}{\sqrt{10}} \) and \( \frac{1}{\sqrt{10}} \) respectively, verifying the consistency of these trigonometric identities with geometric principles.
Right Triangle Trigonometry
Right triangle trigonometry is a fundamental concept when working with angles and sides, especially in exercises like converting polar to rectangular forms. A right triangle consists of two legs and a hypotenuse, with one of the angles being exactly 90 degrees.
  • The side opposite to \( \theta \) gives the measure for \( \sin(\theta) \).
  • The side adjacent to \( \theta \) gives the measure for \( \cos(\theta) \).
In the mathematical scenario presented, we visualize \( \theta = \arctan \left( \frac{1}{3} \right) \) as part of a right triangle. The opposite side of this triangle is 1, and the adjacent side is 3. Using the relation \( \sin^2(\theta) + \cos^2(\theta) = 1 \) and the calculated values for these sides, the hypotenuse is confirmed to be \( \sqrt{10} \). This understanding helps students solve for \( \cos(\theta) \) and \( \sin(\theta) \), thereby connecting the abstract concept of trigonometry to tangible geometrical interpretations.

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Most popular questions from this chapter

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.

In Exercises \(1-20\), plot the set of parametric equations by hand. Be sure to indicate the orientation imparted on the curve by the parametrization. $$ \left\\{\begin{array}{l} x=3 \cos (t) \\ y=3 \sin (t) \end{array} \text { for } 0 \leq t \leq \pi\right. $$

In Exercises \(21-24\), plot the set of parametric equations with the help of a graphing utility. Be sure to indicate the orientation imparted on the curve by the parametrization. $$ \left\\{\begin{array}{l} x=\cos (3 t) \\ y=\sin (4 t) \end{array} \text { for } 0 \leq t \leq 2 \pi\right. $$

The goal of this exercise is to use vectors to describe non-vertical lines in the plane. To that end, consider the line \(y=2 x-4 .\) Let \(\vec{v}_{0}=\langle 0,-4\rangle\) and let \(\vec{s}=\langle 1,2\rangle .\) Let \(t\) be any real number. Show that the vector defined by \(\vec{v}=\vec{v}_{0}+t \vec{s}\), when drawn in standard position, has its terminal point on the line \(y=2 x-4\). (Hint: Show that \(\vec{v}_{0}+t \vec{s}=\langle t, 2 t-4\rangle\) for any real number \(t\).) Now consider the non-vertical line \(y=m x+b\). Repeat the previous analysis with \(\vec{v}_{0}=\langle 0, b\rangle\) and let \(\vec{s}=\langle 1, m\rangle .\) Thus any non-vertical line can be thought of as a collection of terminal points of the vector sum of \(\langle 0, b\rangle\) (the position vector of the \(y\) -intercept) and a scalar multiple of the slope vector \(\vec{s}=\langle 1, m\rangle\).

In Exercises \(1-20\), plot the set of parametric equations by hand. Be sure to indicate the orientation imparted on the curve by the parametrization. $$ \left\\{\begin{array}{l} x=3 \cos (t) \\ y=2 \sin (t)+1 \end{array} \text { for } \frac{\pi}{2} \leq t \leq 2 \pi\right. $$

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