Faculty of Mathematics
Waterloo, Ontario N2L 3G1
Centre for Education in
Mathematics and Computing
Complex Numbers
Problem Set
1. Express each of the following complex numbers in the standard form x + iy, where
x, y ∈ R.
√
√
(a) (1 + i 2) + (1 − i 2) = 2
(e)
(b) (7 + 6i) − (4 − 5i) = 3 + 11i
√
√ √
√
(c) ( 2 − i 3)( 2 + i 3) = 5
√
√
1 2 2
1+i 2
√ =− +
i
(f)
3
3
1−i 2
(d) (3 − i)(1 + i)2 = 2 + 6i
(g)
7−i
=1–i
4 + 3i
1
= −i
i65
2. What are the real and imaginary parts of the following complex numbers?
(a) 4 − 7i Re(z) = 4, Im(z) = −7
√
(b) −10 + 2i
√
Re(z) = −10, Im(z) = 2
(c) −3 Re(z) = −3, Im(z) = 0
(d) (6 − i)2
Re(z) = 35, Im(z) = −12
3. What is the complex conjugate and modulus of each of the following numbers?
(a) 3 − 2i
√
z = 3 + 2i, |z| = 13
(c) −i
z = i, |z| = 1
i
(d) 2 + √
√
(b) 1 + i 5
√
√
z = 1 − i 5, |z| = 6
2
i
3
z = 2 − √ , |z| = √
2
2
1
4. Plot the following points in the complex plane.
(a) z = 1 − i, w = 2 + i, 2z, z + w, and z − 2w
2
(b) z = 2i, z 2 , z 3 , and 1/z
3
5. Find polar coordinates for each of the following points that are given in Cartesian form.
√
(a) (2, 2) = ( 2, π/4)
√
(b) (− 3, 1) = (2, π/6)
6. Each of the following pairs of numbers gives the polar coordinates of a point in the
plane. Express each point in Cartesian coordinates.
√
7 3 7
(b) (7, 7π/6) = (−
,− )
2
2
(a) (1, π/2) = (0, 1)
7. Express each of the following complex numbers in their polar form.
√ (a) −1 − i = 2 · cos(5π/4) + i sin(5π/4)
√
(b) 3 − i = 2 · cos(11π/6) + i sin(11π/6)
(c) 5 = 5 · cos(0) + i sin(0) = 5
8. Express each of the following numbers in the standard form x + iy where x, y ∈ R
(a) cos(3π/2) + i sin(3π/2) = −i
(b)
√
3(cos(4π/3) + sin(4π/3)) = −
√
3 3
− i
2
2
9. Find the modulus and argument of each of the following complex numbers.
(b) (1 + i)30
(a) 2 − 2i
√
π
|z| = 2 2, arg(z) = −
4
|z| = 215 , arg(z) = −
10. Express the following complex numbers in standard form.
(a) (1 − i)6 = 8i
√
(b)
3 i
+
2
2
(c) (1 −
√
√
!25
=
3 i
+
2
2
√
3i)8 = −128 · (1 + i 3)
4
3π
2
11. (a) What is the sum i + i2 + i3 + i4 + i5 + . . . + i2014 + i2015 + i2016 .
First we notice that every in will fall into one of those four categories.
i1 = i5
= i9 = i4k+1
=i
(n has a remainder of 1 when divided by 4)
i2 = i6 = i10 = i4k+2 = −1
(n has a remainder of 2 when divided by 4)
i3 = i7 = i11 = i4k+3
= −i
(n has a remainder of 3 when divided by 4)
=1
(n has a remainder of 0 when divided by 4)
i4 = i8 = i12 = i4k
Moreover, the sum of i4k+1 + i4k+2 + i4k+3 + i4k is zero.
Since the series starts with i4k+1 and that i2016 is of the form i4k , then i + i2 +
i3 + i4 + i5 + . . . + i2014 + i2015 + i2016 = 0
(b) Show that (1 + i2n )(1 + in ) is either 0 or 4 if n is a positive integer. Describe the
values of n that yield the various answers.
First we expand the expression:
(1 + i2n )(1 + in ) = 1 + in + i2n + i3n
Then we consider 4 cases:
i. n = 4k + 1
iii. n = 4k + 3
1 + i4k+1 + i2(4k+1) + i3(4k+1)
1 + i4k+3 + i2(4k+3) + i3(4k+3)
= 1 + i4k+1 + (i4k+1 )2 + (i4k+1 )3
= 1 + (−i) + (−i)2 + (−i)3
= 1 + i + i2 + i3
= 1 + (−i) + (−1) + i
=1+i−1−i
=0
=0
iv. n = 4k
ii. n = 4k + 2
1 + i4k + i2(4k) + i3(4k)
1 + i4k+2 + i2(4k+2) + i3(4k+2)
= 1 + 1 + (1)2 + (1)3
= 1 + (−1) + (−1)2 + (−1)3
=1
=0
5
12. Solve the equation z = z 2 for z ∈ C.
Let z = x + yi, then
z = x − yi
z 2 = (x + yi)2 = x2 + 2xyi − y 2
x − yi = x2 + 2xyi − y 2
Equating the real parts together and the imaginary part together, we get
x = x2 − y 2
−y = 2xy
Using (2)
2xy + y = 0
y(2x + 1) = 0
1
y = 0 or x = −
2
For y = 0
x = x2 − 02
x2 − x = 0
x(x − 1) = 0
x = 1, x = 0
∴ z = 1 or z = 0
1
For x = −
2
1 1 2
− = −
− y2
2
2
1 1
2
y = +
4 2
3
2
y =
4√
3
y=±
2 √
√
−1
3
−1
3
z=
+
i or z =
−
i
2
2
2
2
6
(1)
(2)
13. Use De Moivre’s Theorem to show that
cos 4θ = 8 cos4 θ − 8 cos2 θ + 1
sin 4θ = 4 cos θ(sin θ − 2 sin3 θ)
Hint: Consider (cos θ + i sin θ)4 and recall that (a + b)4 = a4 + 4a3 b + 6a2 b2 + 4ab3 + b4
Consider (cos θ + i sin θ)4 = cos(4θ) + i sin(4θ).
Expand (cos θ + i sin θ)4
(cos θ + i sin θ)4 = cos4 θ + 4 cos2 θi sin θ + 6 cos2 θi2 sin2 θ + 4 cos θi3 sin3 θ + i4 sin4 θ
cos 4θ + i sin 4θ = cos4 θ + 4i cos3 θ sin θ − 6 cos2 θ sin2 θ − 4i cos θ sin3 θ + sin4 θ
Equating the real parts together, we get:
cos 4θ = cos4 θ − 6 cos2 θ sin2 θ + sin4 θ
= cos4 θ − 6 cos2 θ(1 − cos2 θ) + (1 − cos2 θ)2
= cos4 θ − 6 cos2 θ + 6 cos4 θ + 1 − 2 cos2 θ + cos4 θ
= 8 cos4 θ − 8 cos2 θ + 1
Equating the imaginary parts together, we get:
sin 4θ = 4 cos3 θ sin θ − 4 cos θ sin3 θ
= 4cosθ(cos2 θ sin θ − sin3 θ)
= 4 cos θ((1 − sin2 θ) sin θ − sin3 θ)
= 4 cos θ(sin θ − sin3 θ − sin3 θ)
= 4 cos θ(sin θ − 2 sin3 θ)
14. (a) If z = r(cos θ + i sin θ) and w = s(cos φ + i sin φ), prove directly that
r
z
= cos(θ − φ) + i sin(θ − φ).
w
s
z
r(cos θ + i sin θ)
=
w
s(cos φ + i sin θ)
r(cos θ + i sin θ)(cos φ − i sin φ)
=
s(cos φ + i sin φ)(cos φ − i sin φ)
r(cos θ cos φ + i sin θ cos φ − i sin φ cos θ − i2 sin θ sin φ)
=
s(cos2 φ − i2 sin2 φ)
r
= (cos θ cos φ + sin θ cos φ + i(sin θ cos φ − sin φ cos θ))
s
r
= cos(θ − φ)
s
7
√
3 − 3i
√ using this result.
(b) Evaluate √
2 + 6i
√
2 + 6i
q√
√
√
s = ( 2)2 + ( 6)2 = 2 2
√
√
6
π
φ = arctan( √ ) = arctan 3 =
3
2
w=
√
3 − 3i
q√
√
r = ( 3)2 + 32 = 2 3
3
5π
θ = arctan(− √ ) =
3
3
√
5π
5π
z = 2 3(cos
+ i sin )
3
3
z=
√
15. The inverse of z, noted z −1 is a number such that zz −1 = 1.
(a) Show that if z = x + yi is a complex, number, then z −1 =
Let z = x + yi. For z −1 to be the inverse, z −1 z = 1.
zz −1 = (x + yi)
z
.
x2 +y 2
(x − yi)
x2 + y 2
=
x2 − y 2 i2
x2 + y 2
=
x2 + y 2
x2 + y 2
=1
(b) Use part a to show that :
z
w
= zw−1 = z
w
w
Let z = x + yi and w = a + bi. We need to show that
zw
zw
=
ww
(a + bi)(a − bi)
zw
a2 + b 2
w
=z× 2
a + b2
= zw−1
=
8
z
zw
= zw−1 =
w
ww