the multiplicative inverse of x
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.