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Chapter 11: Problem 2

Solve each of the following equations leaving (a) \(z^{3}=-1\) (b) \(z^{3}=1\) (c) \(z^{3}-6 \mathrm{j}=0\) your answers in polar form: (a) \(z^{2}=1+\mathrm{j}\) (b) \(z^{2}=1-\mathrm{j}\) (c) \(z^{3}=-2+3 \mathrm{j}\)

Short Answer

Expert verified
(a) The cube roots of -1 are: $$z_1=\sqrt[3]{1}(\cos(\frac{\pi}{3})+j\sin(\frac{\pi}{3}))$$ $$z_2=\sqrt[3]{1}(\cos(\frac{3\pi}{3})+j\sin(\frac{3\pi}{3}))$$ $$z_3=\sqrt[3]{1}(\cos(\frac{5\pi}{3})+j\sin(\frac{5\pi}{3}))$$ (b) The cube roots of 1 are: $$z_1=\sqrt[3]{1}(\cos(\frac{0}{3})+j\sin(\frac{0}{3}))$$ $$z_2=\sqrt[3]{1}(\cos(\frac{2\pi}{3})+j\sin(\frac{2\pi}{3}))$$ $$z_3=\sqrt[3]{1}(\cos(\frac{4\pi}{3})+j\sin(\frac{4\pi}{3}))$$ (c) The cube roots of 6j are: $$z_1=\sqrt[3]{6}(\cos(\frac{\pi}{6})+j\sin(\frac{\pi}{6}))$$ $$z_2=\sqrt[3]{6}(\cos(\frac{5\pi}{6})+j\sin(\frac{5\pi}{6}))$$ $$z_3=\sqrt[3]{6}(\cos(\frac{9\pi}{6})+j\sin(\frac{9\pi}{6}))$$

Step by step solution

01

Convert -1 to polar form

To find the polar form of -1, we can write it as \(-1+0j\). The magnitude is 1 and the angle is \(\pi\). So, \(-1 =1(\cos\pi + j\sin\pi)\). #Step 2: Apply De Moivre's Theorem#
02

Find the cube roots of -1

To find the cube roots, we will use De Moivre's Theorem: $$z=\sqrt[3]{r}(\cos(\frac{\theta+2k\pi}{n})+j\sin(\frac{\theta+2k\pi}{n}))$$ Here, \(n=3\), \(r=1\), and \(\theta=\pi\). Substituting the values and plugging in \(k=0,1,2\) gives us three cube roots: $$z_1=\sqrt[3]{1}(\cos(\frac{\pi}{3})+j\sin(\frac{\pi}{3}))$$ $$z_2=\sqrt[3]{1}(\cos(\frac{3\pi}{3})+j\sin(\frac{3\pi}{3}))$$ $$z_3=\sqrt[3]{1}(\cos(\frac{5\pi}{3})+j\sin(\frac{5\pi}{3}))$$ #b. Solving for z^3= 1# #Step 1: Convert to polar form#
03

Convert 1 to polar form

To find the polar form of 1, we can write it as \(1+0j\). The magnitude is 1 and the angle is \(0\). So, \(1 =1(\cos0 + j\sin0)\). #Step 2: Apply De Moivre's Theorem#
04

Find the cube roots of 1

Using De Moivre's Theorem with \(n=3\), \(r=1\), and \(\theta=0\), we get three cube roots: $$z_1=\sqrt[3]{1}(\cos(\frac{0}{3})+j\sin(\frac{0}{3}))$$ $$z_2=\sqrt[3]{1}(\cos(\frac{2\pi}{3})+j\sin(\frac{2\pi}{3}))$$ $$z_3=\sqrt[3]{1}(\cos(\frac{4\pi}{3})+j\sin(\frac{4\pi}{3}))$$ #c. Solving for z^3 - 6j= 0# #Step 1: Convert to polar form#
05

Convert 6j to polar form

To find the polar form of 6j, we can write it as \(0+6j\). The magnitude is 6 and the angle is \(\frac{\pi}{2}\). So, \(6j =6(\cos(\frac{\pi}{2}) + j\sin(\frac{\pi}{2}))\). #Step 2: Apply De Moivre's Theorem#
06

Find the cube roots of 6j

Using De Moivre's Theorem with \(n=3\), \(r=6\), and \(\theta=\frac{\pi}{2}\), we get three cube roots: $$z_1=\sqrt[3]{6}(\cos(\frac{\pi}{6})+j\sin(\frac{\pi}{6}))$$ $$z_2=\sqrt[3]{6}(\cos(\frac{5\pi}{6})+j\sin(\frac{5\pi}{6}))$$ $$z_3=\sqrt[3]{6}(\cos(\frac{9\pi}{6})+j\sin(\frac{9\pi}{6}))$$ Note: Square roots for three complex numbers can be computed in a similar way, by finding the necessary polar forms and using De Moivre's Theorem with \(n=2\).

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Most popular questions from this chapter

If \(z=4 \angle \frac{\pi}{6}\) find \(z^{6}\) in polar form.

Simplify the following complex numbers and diagram: show them on an Argand diagram: (a) \(3+3 \mathrm{j}\) (b) \(2=4\) (e) \(-\mathrm{i}\) (f) \(-5=1 \mathrm{j}\) (c) \(-0.5\) (d) \(6 \mathrm{j}\) (a) \(j^{2}\) (b) \(-\mathrm{j}^{2}\) (c) \((-\mathrm{j})^{2}\) (d) \(j^{3}\left(\right.\) e) \(j^{4}\)

Use De Moivre's theorem to show that if \(z=\cos \theta+\mathrm{j} \sin \theta\) then (a) \(z^{n}=\cos n \theta+j \sin n \theta\) (b) \(z^{-n}=\cos n \theta-\mathrm{j} \sin n \theta\) Deduce that $$ z^{n}+\frac{1}{z^{n}}=2 \cos n \theta $$ and $$ z^{n}-\frac{1}{z^{n}}=2 \mathrm{j} \sin n \theta $$

Express each of the following in terms of

Simplify (a) \(-j^{2},(b)(-j)^{2},(c)(-j)^{3}\), (d) \(-j^{3}\).
See all solutions

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