91Ó°ÊÓ

(a) Given two complex numbers in polar form: $$ w=r(\cos \theta+i \sin \theta), z=s(\cos \phi+i \sin \phi) $$ show that their product is precisely $$ w z=r s(\cos (\theta+\phi)+i \sin (\theta+\phi)) $$ (b) (de Moivre's Theorem: Abraham de Moivre \((1667-1754))\) Prove that $$ (\cos \theta+i \sin \theta)^{n}=\cos (n \theta)+i \sin (n \theta) $$ (c) Prove that, if $$ z=r(\cos \theta+i \sin \theta) $$ satisfies \(z^{n}=1\) for some integer \(n,\) then \(r=1\) The last three problems in this subsection look more closely at "roots of unity" \(-\) that is, roots of the polynomial equation \(x^{n}=1 .\) In the real domain, we know that: (i) when \(n\) is odd, the equation \(x^{n}=1\) has exactly one root, namely \(x=1 ;\) and (ii) when \(n\) is even, the equation \(x^{n}=1\) has just two solutions, namely \(x=\pm 1\). In contrast, in the complex domain, there are \(n " n^{\text {th }}\) roots of unity" . Problem \(\mathbf{1 3 0}(\mathrm{c})\) shows that these "roots of unity" all lie on the unit circle, centered at the origin. And if we put \(n \theta=2 k \pi\) in Problem \(130(\mathrm{~b})\) we see that the \(n n^{\text {th }}\) roots of unity include the point \(" 1=\cos 0+i \sin 0\) ", and are then equally spaced around that circle with \(\theta=\frac{2 k \pi}{n}(1 \leqslant k \leqslant n-1),\) and form the vertices of a regular \(n\) -gon.

Step by step solution

01

Polar Form Multiplication Identity

Given two complex numbers in polar form \(w = r(\cos(\theta) + i \sin(\theta))\) and \(z = s(\cos(\phi) + i \sin(\phi))\), their product can be expressed when multiplying the magnitudes \(r\) and \(s\), and using the angle sum identities for trigonometric functions: \[ w z = r s \left( (\cos(\theta) \cos(\phi) - \sin(\theta) \sin(\phi)) + i (\cos(\theta) \sin(\phi) + \sin(\theta) \cos(\phi)) \right) \] Using trigonometric identities: \[ \cos(A + B) = \cos(A)\cos(B) - \sin(A)\sin(B) \] \[ \sin(A + B) = \cos(A)\sin(B) + \sin(A)\cos(B) \] Thus, \[ w z = rs(\cos(\theta + \phi) + i \sin(\theta + \phi)) \] showing that the product of two complex numbers in polar form results in a complex number whose magnitude is the product of the magnitudes and whose angle is the sum of the angles.

02

De Moivre's Theorem Proof

We want to prove that \[ (\cos \theta + i \sin \theta)^{n} = \cos(n\theta) + i \sin(n\theta) \] Using mathematical induction on \(n\):- **Base case**: For \(n=1\), the equation holds as \[ (\cos \theta + i \sin \theta) = \cos \theta + i \sin \theta \]- **Inductive step**: Assume the formula holds for \(n = k\), that is, \[ (\cos \theta + i \sin \theta)^{k} = \cos(k\theta)+ i \sin(k\theta) \]Then for \(n=k+1\), the left-hand side is: \[ (\cos \theta + i \sin \theta)^{k+1} = (\cos \theta + i \sin \theta)^{k} (\cos \theta + i \sin \theta) \]Using the inductive hypothesis, we can substitute:\[ = (\cos(k\theta)+ i \sin(k\theta)) (\cos \theta + i \sin \theta) \]Using the polar form multiplication finding: \[ = (\cos(k\theta + \theta) + i \sin(k\theta + \theta)) \]\[ = (\cos((k+1)\theta) + i \sin((k+1)\theta)) \] Thus, by induction, the formula holds for all integers \(n\).

03

Proving the Magnitude Condition for Unity Roots

Given \(z = r(\cos \theta+i \sin \theta)\) and \(z^{n}=1\), first rewrite \(1\) in polar form as \(1(\cos 0 + i \sin 0)\).Calculating the nth power:\[ z^{n} = (r(\cos \theta + i \sin \theta))^{n} = r^{n}(\cos(n\theta) + i \sin(n\theta)) \]Equating to \(1\):\[ r^{n}(\cos(n\theta) + i \sin(n\theta)) = 1(\cos 0 + i \sin 0) \]If the magnitudes are equal, then:\[ r^{n} = 1 \Rightarrow r=1 \] since \(r\) is real and positive.This shows that if \(z^{n} = 1\), then indeed \(r=1\), confirming that "roots of unity" lie on the unit circle.

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Key Concepts

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

Multiplication of Complex Numbers

Multiplying complex numbers can be easier when they're in polar form. A complex number can be represented as \( w = r(\cos \theta + i \sin \theta) \) and \( z = s(\cos \phi + i \sin \phi) \). To multiply these two, the magnitudes \( r \) and \( s \) are multiplied together, and the angles \( \theta \) and \( \phi \) are added up.
Using the trigonometric identities:
- \( \cos(A + B) = \cos(A)\cos(B) - \sin(A)\sin(B) \)
- \( \sin(A + B) = \cos(A)\sin(B) + \sin(A)\cos(B) \)
we find that when the two polar forms multiply, their resulting form is:
\[ w z = r s(\cos(\theta + \phi) + i \sin(\theta + \phi)) \]
This result shows that the product of two complex numbers in polar form is another complex number, with its magnitude being the product of the two magnitudes, and its "angle" being the sum of the angles.

De Moivre's Theorem

De Moivre's Theorem provides a very handy way to calculate powers of complex numbers in polar form. The theorem states:
\[ (\cos \theta + i \sin \theta)^n = \cos(n\theta) + i \sin(n\theta) \]
This means that raising a complex number (in polar form) to the power of \( n \) simply raises its angle by \( n \) times. A simple way to understand this is through mathematical induction.
**Base case**: For \( n = 1 \), the expression naturally holds as:
\[ (\cos \theta + i \sin \theta) = \cos \theta + i \sin \theta \]
**Inductive step**: We assume it holds for a case when \( n = k \), meaning:
\[ (\cos \theta + i \sin \theta)^k = \cos(k\theta) + i \sin(k\theta) \]
Then for \( n = k + 1 \), we calculate by multiplying:
\( (\cos \theta + i \sin \theta)^{k+1} = (\cos(k\theta) + i \sin(k\theta))(\cos \theta + i \sin \theta) \)
Using the multiplication and angle addition mentioned before, we find that the theorem holds. Thus, De Moivre’s Theorem is a powerful tool for calculations involving powers of complex numbers in polar form.

Roots of Unity

Roots of unity are specific complex numbers that, when raised to a specified power, equal one. In polar form, a complex number \( z = r(\cos \theta + i \sin \theta) \) satisfies \( z^n = 1 \).
To solve for these, we represent 1 as \( 1(\cos 0 + i \sin 0) \), meaning roots of the equation are:
\[ z^n = r^n(\cos(n\theta) + i \sin(n\theta)) = 1 \]
For this equation to hold, \( r^n = 1 \), which means \( r = 1 \), implying that these roots lie on the unit circle. The angles are given by \( n\theta = 2k\pi \), resulting in \( \theta = \frac{2k\pi}{n} \), where \( k \) ranges from 0 to \( n-1 \).
This implies that the \( n \)th roots of unity are evenly distributed around the unit circle, forming the vertices of a regular \( n \)-gon.

Mathematical Induction

Mathematical induction is a proof technique commonly employed in mathematics to verify the truth of an infinite sequence of statements. The process consists of two main parts: the base case and the inductive step.
**Base Case**: One proves that the statement holds for the initial value (usually \( n=1 \)). This verifies that the property is true at the starting point of the sequence.
**Inductive Step**: Here, one assumes that the statement is true for a particular case \( n=k \), called the "induction hypothesis." You then show this implies the statement is also true for \( n=k+1 \).
This technique is often used to prove assertions like De Moivre’s Theorem, where the formula is established for a simple value, and its consistency is demonstrated for all larger values. If both parts succeed, the statement holds universally for all integers following the base case.