# Taking the absolute value preserves multiplication: |z1z2| = |z1||z2| . Triangle inequality |z1 + z2| ≤ |z1|+ z2| .

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.