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Chapter 1: Problem 3

Solve in \(\mathbb{C}\) the equations: a) \(z^{2}+z+1=0\); b) \(z^{3}+1=0\)

Short Answer

Expert verified
The purpose of determining the discriminant in a quadratic equation is to find out whether it has real or complex roots and how many. A positive discriminant indicates two distinct real roots, a negative discriminant indicates two complex conjugate roots, and a zero discriminant indicates one real root (a repeated root or a double root). 2) What theorem is used to find the cube roots of a complex number? De Moivre's theorem is used to find the cube roots of a complex number. This theorem helps to calculate the nth roots of a complex number in polar form. 3) What are the solutions for the quadratic equation \(z^{2}+z+1=0\)? The solutions for the quadratic equation \(z^{2}+z+1=0\) are \(z_{1}=\frac{-1+\sqrt{3}i}{2}\) and \(z_{2}=\frac{-1-\sqrt{3}i}{2}\). 4) What are the solutions for the cubic equation \(z^{3}+1=0\)? The solutions for the cubic equation \(z^{3}+1=0\) are \(z_{1}=\frac{1}{2}+\frac{\sqrt{3}}{2}i\), \(z_{2}=-1\), and \(z_{3}=\frac{1}{2}-\frac{\sqrt{3}}{2}i\).

Step by step solution

01

Determine the discriminant

First, calculate the discriminant \((\Delta)\) of the quadratic equation, which is given by the formula \(\Delta = b^{2} - 4ac\). In this case, \(a=1\), \(b=1\), and \(c=1\). So, \(\Delta = (1)^{2} - 4(1)(1) = 1 - 4 = -3\).
02

Solve for z

Since the discriminant is negative, we know that there are two complex roots. Use the quadratic formula to solve for z: \(z = \frac{-b \pm \sqrt{\Delta}}{2a}\). Plug the values of a, b, and \(\Delta\), which gives: \(z = \frac{-1 \pm \sqrt{-3}}{2}\). The two solutions are: \(z_{1} = \frac{-1+\sqrt{3}i}{2}\) and \(z_{2} = \frac{-1-\sqrt{3}i}{2}\). #b) Cubic Equation: \(z^{3}+1=0\)
03

Rewrite the equation

In this case, we have the equation in the form \(z^{3}+1=0\). Rewrite the equation as \(z^{3}=-1\).
04

Calculate the cube root of -1

We need to find the cube root of -1. We can express -1 as a complex number in polar form: \(-1 = 1\mathrm{cis}(180^\circ)\). The cube root of -1 can be found using De Moivre's theorem, which states that for a complex number in polar form with magnitude r and argument \(\theta\), its nth root is given by: \(\sqrt[n]{r\mathrm{cis}(\theta)} = \sqrt[n]{r}\mathrm{cis}(\frac{\theta+360^\circ k}{n})\), where \(k=0,1,2,\cdots,n-1\). For the cube root of -1, \(n=3\), \(r=1\), and \(\theta=180^\circ\).
05

Apply De Moivre's theorem

Substitute the values of r, \(\theta\), and n into the formula for De Moivre's theorem to find the solutions of the equation. For \(k=0\): \(z_{1}=\sqrt[3]{1}\mathrm{cis}(\frac{180^\circ+360^\circ(0)}{3}) = \mathrm{cis}(60^\circ)\) For \(k=1\): \(z_{2}=\sqrt[3]{1}\mathrm{cis}(\frac{180^\circ+360^\circ(1)}{3}) = \mathrm{cis}(180^\circ)\) For \(k=2\): \(z_{3}=\sqrt[3]{1}\mathrm{cis}(\frac{180^\circ+360^\circ(2)}{3}) = \mathrm{cis}(300^\circ)\)
06

Convert to rectangular form

Finally, convert the solutions from polar form to rectangular form. \(z_{1} = \mathrm{cis}(60^\circ) = \cos(60^\circ) + i\sin(60^\circ) = \frac{1}{2}+\frac{\sqrt{3}}{2}i\) \(z_{2} = \mathrm{cis}(180^\circ) = \cos(180^\circ) + i\sin(180^\circ) = -1\) \(z_{3} = \mathrm{cis}(300^\circ) = \cos(300^\circ) + i\sin(300^\circ) = \frac{1}{2}-\frac{\sqrt{3}}{2}i\) The solutions are: a) \(z_{1}=\frac{-1+\sqrt{3}i}{2}\), \(z_{2}=\frac{-1-\sqrt{3}i}{2}\) b) \(z_{1}=\frac{1}{2}+\frac{\sqrt{3}}{2}i\), \(z_{2}=-1\), \(z_{3}=\frac{1}{2}-\frac{\sqrt{3}}{2}i\)

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Key Concepts

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

Quadratic Equations
Quadratic equations are equations of the form \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are coefficients and \(a eq 0\). The solutions to these equations can be found using the quadratic formula: \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\).

The term inside the square root, \(b^2 - 4ac\), is called the discriminant, represented by the symbol \(\Delta\). The discriminant tells us about the nature of the solutions:
  • If \(\Delta > 0\), the equation has two distinct real roots.
  • If \(\Delta = 0\), the equation has exactly one real root (a repeated root).
  • If \(\Delta < 0\), the equation has two complex roots.
In the given exercise, the quadratic equation \(z^2 + z + 1 = 0\) has a negative discriminant (\(-3\)), indicating that the solutions are complex numbers. By using the quadratic formula, the complex roots are \(z_1 = \frac{-1+\sqrt{3}i}{2}\) and \(z_2 = \frac{-1-\sqrt{3}i}{2}\).
Cubic Equations
Cubic equations are polynomial equations of degree three, generally expressed as \(ax^3 + bx^2 + cx + d = 0\). Unlike quadratic equations, cubic equations can have one real root and two complex conjugate roots, or three real roots (which can include repeated roots).

The process for finding complex roots of cubic equations can be more complex and involves several algebraic techniques or specific formulas depending on the discriminant or its reducibility into factors.

In the exercise given, the equation \(z^3 + 1 = 0\) is already simplified to \(z^3 = -1\). We recognize \(-1\) as a special value easily convertible to its polar form for further solution using De Moivre's Theorem.
Polar Form and De Moivre's Theorem
The polar form of a complex number expresses the number in terms of its magnitude and angle (argument) as \(r \mathrm{cis}(\theta)\), where \(r\) is the magnitude and \(\theta\) is the angle in degrees or radians.

De Moivre's Theorem is a powerful tool for finding powers and roots of complex numbers. It states that for any complex number in polar form, \( [r \mathrm{cis}(\theta)]^n = r^n \mathrm{cis}(n\theta) \).

To find the cube roots of \(-1\), we need to express \(-1\) in polar form: \(-1 = 1 \mathrm{cis}(180^\circ)\).
  • Apply De Moivre's Theorem: Find \(\sqrt[3]{1\mathrm{cis}(180^\circ)}\) by considering the general formula for roots: \(\sqrt[n]{r \mathrm{cis}(\theta)} = \sqrt[n]{r} \mathrm{cis}\left(\frac{\theta + 360^\circ k}{n}\right)\), for \(k = 0, 1, 2, \ldots, n-1\).
  • This results in three cube roots: \(\mathrm{cis}(60^\circ), \mathrm{cis}(180^\circ), \mathrm{cis}(300^\circ)\), shown as \(z_1\), \(z_2\), and \(z_3\).
Complex Roots
Complex roots occur in both quadratic and cubic equations when the discriminant is negative or when higher-degree equations are reduced to simpler forms. These roots are expressed in the form \(a + bi\), where \(i = \sqrt{-1}\).

In quadratic equations, complex roots usually appear in conjugate pairs such as \(z = \frac{-b \pm \, \sqrt{\Delta} i}{2a}\), where \(b^2 - 4ac < 0\).
  • For the quadratic example \(z^2 + z + 1 = 0\), we derived two complex conjugate roots, \(z_1\) and \(z_2\), due to the negative discriminant.
  • For the cubic equation, the roots are derived using polar coordinate conversion and De Moivre's Theorem, yielding complex numbers in rectangular form (after conversion from polar form).
Understanding complex roots deepens our insight into polynomial equations and reveals the broad nature of solutions beyond real numbers.

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Most popular questions from this chapter

Find the geometric image of the complex number \(z\) in each of the following cases: a) \(|z-2|=3\) b) \(|z+i|<1 ; \quad\) c) \(|z-1+2 i|>3\); d) \(|z-2|-|z+2|<2 ; \quad\) e) \(0<\operatorname{Re}(i z)<1 ; \quad\) f) \(-1<\operatorname{Im}(z)<1 ;\) g) \(\operatorname{Re}\left(\frac{z-2}{z-1}\right)=0\) h) \(\frac{1+\bar{z}}{z} \in \mathbb{R}\)

$$ \text { Let } z_{0}=(a, b) \in \mathbb{C} . \text { Find } z \in \mathbb{C} \text { such that } z^{2}=z_{0} \text { . } $$

$$ \text { Find all complex numbers } z \text { such that }|z|=\left|\frac{1}{7}\right| \text { . } $$

Consider \(z \in \mathbb{C}\) with \(\operatorname{Re}(z)>1\). Prove that $$ \left|\frac{1}{z}-\frac{1}{2}\right|<\frac{1}{2} $$

Solve the equations: a) \(z+(-5,7)=(2,-1) ; \quad\) b) \((2,3)+z=(-5,-1)\) c) \(z \cdot(2,3)=(4,5)\) d) \(\frac{z}{(-1,3)}=(3,2)\).
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