# The fundamental Theorem of Algebra.

MATH 4123, HOMEWORK, EXAMS AND SOLUTIONS, FALL 2018

T. PRZEBINDA

Contents

1. Complex numbers. 1 1.1. The field of complex numbers 1 1.2. Multiplication in polar coordinates 3 1.3. Some geometry and some analysis 5 2. The finite Fourier Transform. 7 2.1. The L2 Theory 7 2.2. The L1 Theory 9 3. The Fourier series. 14 3.1. The L1 Theory 14 3.2. Exam 1, due Wednesday 10/3/2018 in class. 17 3.3. The L1 Theory continued 22 3.4. The L2 Theory 23 4. The Fourier Transform on the Schwartz space S(R) 27 4.1. Basic properites of The Fourier Transform 27 4.2. Eigenvectors of Fourier Transform and Quantum Oscillator 30 4.3. The Heisenberg Uncertainty Principle 35 5. Exam 2, due Monday 11/26/2018 in class. 36 References 38

1. Complex numbers.

1.1. The field of complex numbers.

C ∈ z = x+ iy ↔ (x, y) ∈ R2 .

The addition of complex numbers is defined coordinate-wise:

(x1 + iy1) + (x2 + iy2) = (x1 + x2) + i(y1 + y2)

and multiplication by

(x1 + iy1)(x2 + iy2) = x1x2 − y1y2 + i(x1y2 + x2y1) .

Date: November 21, 2018.

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2 T. PRZEBINDA

Then i2 = −1 and the multiplicative inverse of x+ iy is

(x+ iy)−1 = x

x2 + y2 − i y

x2 + y2 .

With these operations C is a field, i.e. behaves like the field R of real numbers.

Problem 1. Check that

(x+ iy)

( x

x2 + y2 − i y

x2 + y2

) = 1 .

The left hand side is equal to

(x+ iy)(x− iy) x2 + y2

= x2 + y2

x2 + y2 = 1 .

In problems 2 and 3 we realize complex numbers as matricesof size two with real entries.

Problem 2. Let

I =

( 1 0 0 1

) and J =

( 0 1 −1 0

) .

Show that for any real numbers x1, x2, y1 and y2,

(x1I + y1J) (x2I + y2J) = (x1x2 − y1y2)I + (x1y2 + x2y1)J. In particular

(x1I + y1J) (x2I + y2J) = (x2I + y2J) (x1I + y1J)

and J2 = −I.

We notice first that J2 = −I.

Then

(x1I + y1J) (x2I + y2J) = x1Ix2I + x1Iy2J + y1Jx2I + y1Jy2J

= x1x2I + x1y2J + y1Jx2 − y1y2I = (x1x2 − y1y2)I + (x1y2 + x2y1)J.

Problem 3. With the notation of Problem 2 show that the matrix

xI + yJ

is invertible if and only if x2 + y2 6= 0. Also, show that

(xI + yJ)−1 = x

x2 + y2 I − y

x2 + y2 J.

MATH 4123, HOMEWORK, EXAMS AND SOLUTIONS, FALL 2018 3

If x2 + y2 6= 0, then the matrix on the right hand side is well defined and we see from the formula of previous problem that

(xI + yJ)

( x

x2 + y2 I − y

x2 + y2 J

) = I.

Thus the matrix xI + yJ is invertible. If x2 + y2 = 0, then xI + yJ = 0 is not an invertible matrix.

Complex conjugation x+ iy = x− iy

is an automorphism of C. In other words the conjugation of the sum is the sum of conjugates

(x1 + iy1) + (x2 + iy2) = x1 + iy1 + x2 + iy2 and the conjugation of the product is the product of conjugates

(x1 + iy1)(x2 + iy2) = x1 + iy1 x2 + iy2 .

Also, {z ∈ C; z = z} = R .

Problem 4. Let a, b, c be complex (or real) numbers. Show that if z solves the equation

az2 + bz + c = 0

then z solves the equation a(z)2 + b(z) + c = 0.

This is clear because az2 + bz + c = a(z)2 + b(z) + c.