Chapter 8 Complex Numbers
MATH 135 CLASSNOTE FOR MAKEUP CLASS ON NOVEMBER 16, 2010 WENTANG KUO
Chapter 8
Complex Numbers
We recall the Quadradic formula: if a, b, c ∈ R where a 6= 0, then the solution of ax2 + bx + c = 0 is
√
b2 − 4ac . 2a The solutions are if b2 − 4ac ≥ 0. What if b2 − 4ac < 0? x=
8.2
−b ±
Complex numbers
Definition Let the symbol i be a square root of −1, i.e.,i2 = −1. A complex number z is a number of the form z = x + iy where x, y ∈ R (Cartesian form). We call x the real part of z, denoted by Re(z) and y the imaginary part of z, denoted it by Im(z). Note that x ∈ R if and only if y = 0. The set of all complex numbers is denoted by C. The addition and multiplication on C are defined as follows: let z = x + iy and w = u + iv. Then z + w = (x + iy) + (u + iv) = (x + u) + i(y + v) and z · w = (x + iy) · (u + iv) = xu + iyu + xiv + i2 yv = (xu − yv) + i(yu + xv). Example Let z = 2 + i and w = 4 − i. Find Im(z + w) and Re(z 2 w). We have z + w = (2 + 4) + i(1 + (−1)) = 6 + 0 · i = 6. Thus, Im(z + w) = 0. Also, z 2 w = (2 + i)2 (4 − i) = (22 + 4i + i2 )(4 − i) = (3 + 4i)(4 − i) = 16 + 13i. Thus, Re(z 2 w) = 16. √ Example Find z ∈ C such that z 2 = 1 + 2 2i. Write z = x + iy with x, y ∈ R. Then
√ z 2 = (x + iy)2 = x2 + 2xyi + i2 y 2 = (x2 − y 2 ) + 2xyi = 1 + 2 2i. √ √ √ Thus, we have x2 − y 2 = 1 and 2xy = 2 2, i.e., xy = 2. Write y = x2 . We have √ 2 2 2 2 2 2 x −y =x − = x2 − 2 = 1. x x Date: November 15, 2010.
1
2
WENTANG KUO
Multiplying by x2 , we get x4 − 2 = x2 ,
(x2 − 2)(x2 + 1) = 0. √ √ √ Hence, x2 =√2 or −1. Since x√∈ R, x2 ≥ 0. Hence, x2 = 2, i.e., x = ± 2. Also, y = 2/± 2 = ±1. Hence, z = 2 + i or z = − 2 − i.
8.3
i.e., x4 − x2 − 2 = 0,
The complex plane
&
8.4
i.e.,
Properties of complex numbers
The complex plane A way to visualize C and its operations. Idea Let x-axis be the real axis and y-axis the imaginary axis. We associate z = x + iy to the point (x, y). imaginary 6 y
1s z = x + iy = (x, y) -
x real Example Let z = x + iy and w = u + iv. Then z + w = (x + u) + i(y + v). 6
y+v
z+w > w v s y 1s z u
x
-
x+u
Addition can be visualized as the parallelogram law for adding vectors. Given z = x + iy ∈ C, we have (x + iy) · (x − iy) = x2 + y 2 ∈ R. (Thus, by multiplying (x + iy) by (x − iy), we get back a real number) Definition The complex conjugate of z = x + iy is z¯ = x − iy. 6 1s z = x + iy PP PP x PP PP qs z¯ = x − iy P −y
y
MATH 135 CLASSNOTE FOR MAKEUP CLASS ON NOVEMBER 16, 2010
Conjugation is just the reflection with respect to the real axis. Definition The modulus of z = x + iy is p |z| = x2 + y 2 . Note that the distance from (0, 0) to (x, y) is p x2 + y 2 = |z|.
6
r=
p x2 + y 2 1s z = x + iy y
-
x Thus, |z| is the distance from 0 to z. Remark (1) We have z · z¯ = |z|2 . (2) Note that if y = 0, i.e., z = x ∈ R, then |z| =
√
x2 = |x|,
i.e., for real numbers, modulus is just the absolute value. Example Find the inverse of z = x + iy ∈ C, where x and y are not both 0. We have 1 1 x − iy x − iy z¯ 1 = · = 2 = 2. z −1 = = 2 z x + iy x + iy x − iy x +y |z| Example Find z, w ∈ C such that z + w = 7 — (1) 2iz + 7w = 9
— (2).
From (1), we have z = 7 − w. Put it into (2). We get 2i(7 − w) + 7w = 9,
i.e.,
(7 − 2i)w = 9 − 14i.
Thus, w=
9 − 14i 9 − 14i 7 + 2i 63 − 98i + 18i − 28i2 (63 + 28) − 80i 91 − 80i = · = = = . 2 2 2 2 7 − 2i 7 − 2i 7 + 2i 7 +2 7 +2 53
Also, z =7−w =7− Properties of Conjugation 8.42 If z, w ∈ C, we have (1) z + w = z¯ + w. ¯ (2) z · w = z¯ · w. ¯ (3) z¯ = z. (4) z · z¯ = |z|2 .
91 − 80i 371 − (91 − 80i) 280 + 80i = = . 53 53 53
3
4
WENTANG KUO
(5) z + z¯ = 2 Re(z). (6) z − z¯ = 2i Im(z). Proof. (1) and (2) are easy and we have shown (4). Let z = x + iy. Then z¯ = x − iy. (3) We have z¯ = x − iy = x + iy = z. (5) We have z + z¯ = (x + iy) + (x − iy) = 2x = 2 Re(z). (6) We have z − z¯ = (x + iy) − (x − iy) = 2iy = 2i Im(z). Properties of Modulus 8.44 (1) |z| = 0 if and only if z = 0. (2) |¯ z | = |z|. (3) |zw| = |z||w|. (4) |z + w| ≤ |z| + |w| (triangle Inequality). Proof. Let z = x + iy. (1) |z| = 0 ⇐⇒
p x2 + y 2 = 0 ⇐⇒ x2 + y 2 = 0 ⇐⇒ x = 0 = y, i.e., z = 0.
(2) |¯ z | = |x − iy| =
p
(x2 + (−y)2 =
p x2 + y 2 = |z|.
(3) Since |zw| and |z||w| are non-negative real number, it suffices to prove |zw|2 = (|z||w|)2 . We have |zw|2 = (zw)(zw)
(by Proposition 8.42(4))
= (zw)(¯ z w) ¯
(by Proposition 8.42(2))
= (z z¯)(ww) ¯ = |z|2 |w|2
(by Proposition 8.42(4))
= (|z||w|)2 . It follows that |zw| = |z||w|. (4) 6
z+w > w s |z+w| |w| 1s z |z|
-
MATH 135 CLASSNOTE FOR MAKEUP CLASS ON NOVEMBER 16, 2010
5
We have the sum of two sides of a triangle ≥ the third side. Since |z + w| and (|z| + |w|) are ≥ 0, it suffices to prove |z + w|2 ≤ (|z| + |w|)2 . |z + w|2 = (z + w) · (z + w)
(by Proposition 8.42(4))
= (z + w) · (¯ z + w) ¯ (by Proposition 8.42(1)) = z z¯ + w¯ z + zw ¯ + ww ¯ = |z|2 + |w|2 + w¯ z + zw ¯ = |z|2 + |w|2 + w¯ z + w¯ z (by Proposition 8.42(3)) = |z|2 + |w|2 + 2Re(w¯ z ). (by Proposition 8.42(5)) Also, we have (|z| + |w|)2 = |z|2 + 2|z||w| + |w|2 = |z|2 + |w|2 + 2|¯ z ||w| (by Proposition 8.42(2)) = |z|2 + |w|2 + 2|w¯ z| Note that for any t = a + ib ∈ C, |t| =
(by Proposition 8.44(3))
p √ a2 + b2 ≥ a2 = |a| ≥ Re (t).
Thus, Re(w¯ z ) ≤ |w¯ z |, which implies that |z + w|2 ≤ (|z| + |w|)2 . Thus, we have |z + w| ≤ |z| + |w|. Example Draw the picture of the set z ∈ C | 2 ≤ |z + 1| < 3 . We recall that |z| is the distance between 0 and z. In general, for a ∈ C, |z − a| is the distance between z and a. Since |z + 1| = |z − (−1)|, 2 ≤ |z + 1| < 3 states that the distance between z and (−1) is ≥ 2 and < 3. Note that |z − (−1)| = 2 is the circle of radius 2 whose center is (−1). y 6
q
q
q -
(−1,0)
(1,0)(2,0)
x
Example Draw the picture of the set z ∈ C | |z − 1| = |z − i| .
6
WENTANG KUO
We have |z − 1| = |z − i| when z is equidistant from 1 and i. Thus, z lies in a line that bisects and is perpendicular to the line segment from 1 to i. y 6
i q q
x -
1
8.5
Polar representation
Polar Coordinates Another way to visualize C in terms of real numbers. s z = x + iy
6
r
θ -
o
Idea Using r and θ to identity z, where r = |z| =
p x2 + y 2 = the modulus of z,
and θ = angle (measured in radians) between the x-axis and the line oz = the argument of z. We call (r, θ) the polar coordinates of z. Note that if z = (r, θ) = x + iy, we have x = r cos θ,
and y = r sin θ.
Then the polar form of z is z = r(cos θ + i sin θ). Example Express z = 1 + i in terms of polar coordinates and polar form.
MATH 135 CLASSNOTE FOR MAKEUP CLASS ON NOVEMBER 16, 2010
sz = 1 + i
6
r
1
θ = π/4 -
1 Hence, we have r=
p √ 11 + 1 2 = 2
and θ = π/4.
Hence,
√ √ 1 + i = ( 2, π/4) and 1 + i = 2(cos π/2 + i sin π/4). Example Express z = −2 = −2 + 0i in terms of polar coordinates. s
−2
6 θ = 3π -
θ=π r=2
Hence, −2 = (2, π). Note that (2, 3π) represents the same point, so does (2, 5π). Remark In general, since the angel of a full circle is 2π, (r, θ) = (r, θ + 2nπ),
∀ n ∈ Z,
i.e., r(cos θ + i sin θ) = r[cos(θ + 2nπ) + i sin(θ + 2nπ)], Converting between two coordinates
∀ n ∈ Z.
6
r θ
1s z = x + iy
y
-
x From polar coordinates to standard forms: Given r, θ, we have x = r cos θ, y = r sin θ. 11π Example Convert (r, θ) = (2, 6 ) into its standard form. √ 11π 3√ 11π 1 x = 2 cos =2· 3 and y = 2 sin = 2 · − = −1. 6 2 6 2 √ 11π Hence, (2, 6 ) = 3 − i. From standard forms to polar coordinates: Given x, y, we have p y r = x2 + y 2 and tan θ = . x
7
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WENTANG KUO
To get θ, we compute tan−1 xy . Warning The range of tan−1 is −π/2 < tan−1 < π/2. However, out argument is from −π/2 to 3π/2. Hence, θ = tan−1 xy or θ = π + tan−1 xy , depending on the signs of x and y. √ Example Convert √ −i, 3 + 3i, and −3 + 3i into polar forms. We have | − i| = 02 + 12 = 1. Also, the argument is 3π/2. Hence, −i = 1(cos 3π/2 + i sin 3π/2). We have |3 + 3i| =
√
9+9=
√
18. Also, the argument is tan−1
3 1 = tan−1 = π/4. 3 1
Since both x, y > 0, we have 0 < θ < π/2. Hence, 3 + 3i = We have | − 3 +
√
3i| =
√
9+3=
√
√
18 (cos π/4 + i sin π/4).
12. Also, the argument is √
−1
tan
3 1 = tan−1 √ = −π/6. −3 − 3
Since x < 0, y > 0, π/2 < θ < π. Hence, θ = −π/6 + π = 5π/6. It follows that −3 +
√ √ 3i = 12 (cos 5π/6 + i sin 5π/6).
Theorem 8.53** Let z1 = r1 (cos θ1 + i sin θ1 ), z2 = r2 (cos θ2 + i sin θ2 ) ∈ C. Then z1 · z2 = r1 · r2 [cos(θ1 + θ2 ) + i sin(θ1 + θ2 )]. Proof. We recall that cos(θ1 + θ2 ) = cos θ1 cos θ2 − sin θ1 sin θ2 and sin(θ1 + θ2 ) = sin θ1 cos θ2 + sin θ2 cos θ1 . Thus, z1 z2 = r1 (cos θ1 + i sin θ1 ) · r2 (cos θ2 + i sin θ2 ) = r1 r2 [cos θ1 cos θ2 − sin θ1 sin θ2 + i(sin θ1 cos θ2 + sin θ2 cos θ1 )] = r1 r2 [cos(θ1 + θ2 ) + i sin(θ1 + θ2 )].
MATH 135 CLASSNOTE FOR MAKEUP CLASS ON NOVEMBER 16, 2010
9
z1 z2 s
z1 s 6
z2
s
θ1 + θ2
θ2 θ1
-
Hence, to multiply two complex numbers, it is equivalent to multiply their modulus and add their angles. Department of Pure Mathematics, University of Waterloo, Waterloo, Ontario, Canada E-mail address: [email protected]
Chapter 8
Complex Numbers
We recall the Quadradic formula: if a, b, c ∈ R where a 6= 0, then the solution of ax2 + bx + c = 0 is
√
b2 − 4ac . 2a The solutions are if b2 − 4ac ≥ 0. What if b2 − 4ac < 0? x=
8.2
−b ±
Complex numbers
Definition Let the symbol i be a square root of −1, i.e.,i2 = −1. A complex number z is a number of the form z = x + iy where x, y ∈ R (Cartesian form). We call x the real part of z, denoted by Re(z) and y the imaginary part of z, denoted it by Im(z). Note that x ∈ R if and only if y = 0. The set of all complex numbers is denoted by C. The addition and multiplication on C are defined as follows: let z = x + iy and w = u + iv. Then z + w = (x + iy) + (u + iv) = (x + u) + i(y + v) and z · w = (x + iy) · (u + iv) = xu + iyu + xiv + i2 yv = (xu − yv) + i(yu + xv). Example Let z = 2 + i and w = 4 − i. Find Im(z + w) and Re(z 2 w). We have z + w = (2 + 4) + i(1 + (−1)) = 6 + 0 · i = 6. Thus, Im(z + w) = 0. Also, z 2 w = (2 + i)2 (4 − i) = (22 + 4i + i2 )(4 − i) = (3 + 4i)(4 − i) = 16 + 13i. Thus, Re(z 2 w) = 16. √ Example Find z ∈ C such that z 2 = 1 + 2 2i. Write z = x + iy with x, y ∈ R. Then
√ z 2 = (x + iy)2 = x2 + 2xyi + i2 y 2 = (x2 − y 2 ) + 2xyi = 1 + 2 2i. √ √ √ Thus, we have x2 − y 2 = 1 and 2xy = 2 2, i.e., xy = 2. Write y = x2 . We have √ 2 2 2 2 2 2 x −y =x − = x2 − 2 = 1. x x Date: November 15, 2010.
1
2
WENTANG KUO
Multiplying by x2 , we get x4 − 2 = x2 ,
(x2 − 2)(x2 + 1) = 0. √ √ √ Hence, x2 =√2 or −1. Since x√∈ R, x2 ≥ 0. Hence, x2 = 2, i.e., x = ± 2. Also, y = 2/± 2 = ±1. Hence, z = 2 + i or z = − 2 − i.
8.3
i.e., x4 − x2 − 2 = 0,
The complex plane
&
8.4
i.e.,
Properties of complex numbers
The complex plane A way to visualize C and its operations. Idea Let x-axis be the real axis and y-axis the imaginary axis. We associate z = x + iy to the point (x, y). imaginary 6 y
1s z = x + iy = (x, y) -
x real Example Let z = x + iy and w = u + iv. Then z + w = (x + u) + i(y + v). 6
y+v
z+w > w v s y 1s z u
x
-
x+u
Addition can be visualized as the parallelogram law for adding vectors. Given z = x + iy ∈ C, we have (x + iy) · (x − iy) = x2 + y 2 ∈ R. (Thus, by multiplying (x + iy) by (x − iy), we get back a real number) Definition The complex conjugate of z = x + iy is z¯ = x − iy. 6 1s z = x + iy PP PP x PP PP qs z¯ = x − iy P −y
y
MATH 135 CLASSNOTE FOR MAKEUP CLASS ON NOVEMBER 16, 2010
Conjugation is just the reflection with respect to the real axis. Definition The modulus of z = x + iy is p |z| = x2 + y 2 . Note that the distance from (0, 0) to (x, y) is p x2 + y 2 = |z|.
6
r=
p x2 + y 2 1s z = x + iy y
-
x Thus, |z| is the distance from 0 to z. Remark (1) We have z · z¯ = |z|2 . (2) Note that if y = 0, i.e., z = x ∈ R, then |z| =
√
x2 = |x|,
i.e., for real numbers, modulus is just the absolute value. Example Find the inverse of z = x + iy ∈ C, where x and y are not both 0. We have 1 1 x − iy x − iy z¯ 1 = · = 2 = 2. z −1 = = 2 z x + iy x + iy x − iy x +y |z| Example Find z, w ∈ C such that z + w = 7 — (1) 2iz + 7w = 9
— (2).
From (1), we have z = 7 − w. Put it into (2). We get 2i(7 − w) + 7w = 9,
i.e.,
(7 − 2i)w = 9 − 14i.
Thus, w=
9 − 14i 9 − 14i 7 + 2i 63 − 98i + 18i − 28i2 (63 + 28) − 80i 91 − 80i = · = = = . 2 2 2 2 7 − 2i 7 − 2i 7 + 2i 7 +2 7 +2 53
Also, z =7−w =7− Properties of Conjugation 8.42 If z, w ∈ C, we have (1) z + w = z¯ + w. ¯ (2) z · w = z¯ · w. ¯ (3) z¯ = z. (4) z · z¯ = |z|2 .
91 − 80i 371 − (91 − 80i) 280 + 80i = = . 53 53 53
3
4
WENTANG KUO
(5) z + z¯ = 2 Re(z). (6) z − z¯ = 2i Im(z). Proof. (1) and (2) are easy and we have shown (4). Let z = x + iy. Then z¯ = x − iy. (3) We have z¯ = x − iy = x + iy = z. (5) We have z + z¯ = (x + iy) + (x − iy) = 2x = 2 Re(z). (6) We have z − z¯ = (x + iy) − (x − iy) = 2iy = 2i Im(z). Properties of Modulus 8.44 (1) |z| = 0 if and only if z = 0. (2) |¯ z | = |z|. (3) |zw| = |z||w|. (4) |z + w| ≤ |z| + |w| (triangle Inequality). Proof. Let z = x + iy. (1) |z| = 0 ⇐⇒
p x2 + y 2 = 0 ⇐⇒ x2 + y 2 = 0 ⇐⇒ x = 0 = y, i.e., z = 0.
(2) |¯ z | = |x − iy| =
p
(x2 + (−y)2 =
p x2 + y 2 = |z|.
(3) Since |zw| and |z||w| are non-negative real number, it suffices to prove |zw|2 = (|z||w|)2 . We have |zw|2 = (zw)(zw)
(by Proposition 8.42(4))
= (zw)(¯ z w) ¯
(by Proposition 8.42(2))
= (z z¯)(ww) ¯ = |z|2 |w|2
(by Proposition 8.42(4))
= (|z||w|)2 . It follows that |zw| = |z||w|. (4) 6
z+w > w s |z+w| |w| 1s z |z|
-
MATH 135 CLASSNOTE FOR MAKEUP CLASS ON NOVEMBER 16, 2010
5
We have the sum of two sides of a triangle ≥ the third side. Since |z + w| and (|z| + |w|) are ≥ 0, it suffices to prove |z + w|2 ≤ (|z| + |w|)2 . |z + w|2 = (z + w) · (z + w)
(by Proposition 8.42(4))
= (z + w) · (¯ z + w) ¯ (by Proposition 8.42(1)) = z z¯ + w¯ z + zw ¯ + ww ¯ = |z|2 + |w|2 + w¯ z + zw ¯ = |z|2 + |w|2 + w¯ z + w¯ z (by Proposition 8.42(3)) = |z|2 + |w|2 + 2Re(w¯ z ). (by Proposition 8.42(5)) Also, we have (|z| + |w|)2 = |z|2 + 2|z||w| + |w|2 = |z|2 + |w|2 + 2|¯ z ||w| (by Proposition 8.42(2)) = |z|2 + |w|2 + 2|w¯ z| Note that for any t = a + ib ∈ C, |t| =
(by Proposition 8.44(3))
p √ a2 + b2 ≥ a2 = |a| ≥ Re (t).
Thus, Re(w¯ z ) ≤ |w¯ z |, which implies that |z + w|2 ≤ (|z| + |w|)2 . Thus, we have |z + w| ≤ |z| + |w|. Example Draw the picture of the set z ∈ C | 2 ≤ |z + 1| < 3 . We recall that |z| is the distance between 0 and z. In general, for a ∈ C, |z − a| is the distance between z and a. Since |z + 1| = |z − (−1)|, 2 ≤ |z + 1| < 3 states that the distance between z and (−1) is ≥ 2 and < 3. Note that |z − (−1)| = 2 is the circle of radius 2 whose center is (−1). y 6
q
q
q -
(−1,0)
(1,0)(2,0)
x
Example Draw the picture of the set z ∈ C | |z − 1| = |z − i| .
6
WENTANG KUO
We have |z − 1| = |z − i| when z is equidistant from 1 and i. Thus, z lies in a line that bisects and is perpendicular to the line segment from 1 to i. y 6
i q q
x -
1
8.5
Polar representation
Polar Coordinates Another way to visualize C in terms of real numbers. s z = x + iy
6
r
θ -
o
Idea Using r and θ to identity z, where r = |z| =
p x2 + y 2 = the modulus of z,
and θ = angle (measured in radians) between the x-axis and the line oz = the argument of z. We call (r, θ) the polar coordinates of z. Note that if z = (r, θ) = x + iy, we have x = r cos θ,
and y = r sin θ.
Then the polar form of z is z = r(cos θ + i sin θ). Example Express z = 1 + i in terms of polar coordinates and polar form.
MATH 135 CLASSNOTE FOR MAKEUP CLASS ON NOVEMBER 16, 2010
sz = 1 + i
6
r
1
θ = π/4 -
1 Hence, we have r=
p √ 11 + 1 2 = 2
and θ = π/4.
Hence,
√ √ 1 + i = ( 2, π/4) and 1 + i = 2(cos π/2 + i sin π/4). Example Express z = −2 = −2 + 0i in terms of polar coordinates. s
−2
6 θ = 3π -
θ=π r=2
Hence, −2 = (2, π). Note that (2, 3π) represents the same point, so does (2, 5π). Remark In general, since the angel of a full circle is 2π, (r, θ) = (r, θ + 2nπ),
∀ n ∈ Z,
i.e., r(cos θ + i sin θ) = r[cos(θ + 2nπ) + i sin(θ + 2nπ)], Converting between two coordinates
∀ n ∈ Z.
6
r θ
1s z = x + iy
y
-
x From polar coordinates to standard forms: Given r, θ, we have x = r cos θ, y = r sin θ. 11π Example Convert (r, θ) = (2, 6 ) into its standard form. √ 11π 3√ 11π 1 x = 2 cos =2· 3 and y = 2 sin = 2 · − = −1. 6 2 6 2 √ 11π Hence, (2, 6 ) = 3 − i. From standard forms to polar coordinates: Given x, y, we have p y r = x2 + y 2 and tan θ = . x
7
8
WENTANG KUO
To get θ, we compute tan−1 xy . Warning The range of tan−1 is −π/2 < tan−1 < π/2. However, out argument is from −π/2 to 3π/2. Hence, θ = tan−1 xy or θ = π + tan−1 xy , depending on the signs of x and y. √ Example Convert √ −i, 3 + 3i, and −3 + 3i into polar forms. We have | − i| = 02 + 12 = 1. Also, the argument is 3π/2. Hence, −i = 1(cos 3π/2 + i sin 3π/2). We have |3 + 3i| =
√
9+9=
√
18. Also, the argument is tan−1
3 1 = tan−1 = π/4. 3 1
Since both x, y > 0, we have 0 < θ < π/2. Hence, 3 + 3i = We have | − 3 +
√
3i| =
√
9+3=
√
√
18 (cos π/4 + i sin π/4).
12. Also, the argument is √
−1
tan
3 1 = tan−1 √ = −π/6. −3 − 3
Since x < 0, y > 0, π/2 < θ < π. Hence, θ = −π/6 + π = 5π/6. It follows that −3 +
√ √ 3i = 12 (cos 5π/6 + i sin 5π/6).
Theorem 8.53** Let z1 = r1 (cos θ1 + i sin θ1 ), z2 = r2 (cos θ2 + i sin θ2 ) ∈ C. Then z1 · z2 = r1 · r2 [cos(θ1 + θ2 ) + i sin(θ1 + θ2 )]. Proof. We recall that cos(θ1 + θ2 ) = cos θ1 cos θ2 − sin θ1 sin θ2 and sin(θ1 + θ2 ) = sin θ1 cos θ2 + sin θ2 cos θ1 . Thus, z1 z2 = r1 (cos θ1 + i sin θ1 ) · r2 (cos θ2 + i sin θ2 ) = r1 r2 [cos θ1 cos θ2 − sin θ1 sin θ2 + i(sin θ1 cos θ2 + sin θ2 cos θ1 )] = r1 r2 [cos(θ1 + θ2 ) + i sin(θ1 + θ2 )].
MATH 135 CLASSNOTE FOR MAKEUP CLASS ON NOVEMBER 16, 2010
9
z1 z2 s
z1 s 6
z2
s
θ1 + θ2
θ2 θ1
-
Hence, to multiply two complex numbers, it is equivalent to multiply their modulus and add their angles. Department of Pure Mathematics, University of Waterloo, Waterloo, Ontario, Canada E-mail address: [email protected]
