The left hand side is equal to (x+ iy)(x− iy) x2 + y2 = x2 + y2 x2 + y2 = 1 .

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.

1

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.