Complex eigenvaules
1. (warm-up)
(a) Plot 1 + i and 3 − 2i in the complex plane. What’s |1 + i| and what is |3 − 2i|?
(b) Plot 4eπi/3 in the complex plane using Euler’s formula eiθ = cos θ + i sin θ.
(c) Simplify (1 + i) + (3 − 2i)2
(d) Simplify (1 + i)(3 − 2i)
Solution. In class.
0 −1
2. Let A =
. Find the eigenvaules and corresponding eigenvectors.
1
0
2
Solution. The characteristic polynomial is λ + 1, which has roots ±i.
1
is an eigenvector for −i.
and
i
3. Find the roots of the characteristic polynomial of
2
−2
1
.
4
Solution. The characteristic polynomial is λ2 − 6λ + 10, which has roots
4. Let z = a + bi, and
√
1
is an eigenvector for i,
−i
√
6± −4
2
= 3 ± i.
a2 + b2 = r, then z = reiθ = r(cos θ + i sin θ). Moreover,
z n = rn (cos θ + i sin θ)n
z n = rn eiθn = rn (cos(nθ) + i sin(nθ)).
So, we know
(cos θ + i sin θ)n = cos(nθ) + i sin(nθ).
This is called De Moivre’s formula.
√
√
(a) Express ( 3 − i)65 in the form a + bi. (Hint: Start by writing 3 − i in the form reiθ .)
√
√
11πi
715πi
Solution. 3 − i = 2e 6 , so ( 3 − i)65 = 265 e 6 = 265 ei7π/6
(b) Show that cos(3θ) = 4 cos3 θ − 3 cos θ
Solution. By De Moivre’s formula, we know (cos θ + i sin θ)3 = cos(3θ) + i sin(3θ).
LHS: (cos θ + i sin θ)3 = (4 cos3 θ − 3 cos θ) + i(3 sin θ − 4 sin3 θ)
RHS: cos(3θ) + i sin(3θ)
The real parts and imaginary parts in the LHS and RHS should be the same. Hence
cos(3θ) = 4 cos3 θ − 3 cos θ and sin(3θ) = 3 sin θ − 4 sin3 θ
5. (a) Let z =
√
3 + 3i. Compute |z|
1
(b) Express z =
√
3 + 3i in the form z = reiθ , where r = |z|. What is log z?
Define log z = log |z| + iarg(z), here arg(z) = θ and 0 ≤ θ ≤ 2π.
√
Solution. |z| = 2 3.
r
3
θ
√
3
To express z in polar
looking for the r and θ in√the picture. Using the Pythagorean
√ coordinates, we are √
Theorem, r2 = ( 3)2 + 32 = 12, so r = 2 3. Also, tan θ = √33 = 3, so θ is either π3 or 4π
. Since z is
√ 3 iπ/3
π
. Hence
in the first quadrant of the complex plane, it must be the case that θ = 3 . So, z = 2 3e
√
log z = log(2 3) + iπ/3 .
6. Compute the following:
(a) log(3i)
Solution. 3i = 3(cos π/2 + i sin π/2), so log(3i) = log 3 + iπ/2
(b) (−1)i
b
Hint: use ab = ea log b (Oops, this is a typo. It should be ab = eb log a , remember ab = elog a =
eb log a )
Solution. (−1)i = ei log(−1) and log(−1) = log 1 + iπ = iπ. So (−1)i = ei(iπ) = e−π .

0
0

7. Find the characteristic polynomial for 
0
0
1
1
0
0
0
0
0
1
0
0
0
0
0
1
0
0

0
0

0
. What are the eigenvaules?
1
0
Solution. The characteristic polynomial is λ5 −1. To find the roots, consider λ5 = 1 = ek2πi , where k is
any integer. Since we know there are 5 complex roots (Fundamental theorem of algebra: A polynomial
of degree n has exactly n complex roots), the roots must be λk = ek2πi/5 , and k = 0, 1, 2, 3, 4.
8. (T/F)
(a) If λ1 is an eigenvalue for A, then λ1 + 1 is an eigenvaule for A + I
Solution. True. If A~v = λ~v then (A + I)~v = A~v + ~v = (λ + 1)~v .
(b) If λ1 is an eigenvalue for A, and λ2 is an eigenvalue for B, then λ1 + λ2 is an eigenvaule for A + B
Solution. False. The eigenvectors might not be the same.
2