A complex number is z = a + b i z = a + bi z = a + bi where a , b ∈ R a, b \in \mathbb{R} a , b ∈ R and i 2 = − 1 i^2 = -1 i 2 = − 1 . We call a = R e ( z ) a = \mathrm{Re}(z) a = Re ( z ) the real part and b = I m ( z ) b = \mathrm{Im}(z) b = Im ( z ) the imaginary part.
Arithmetic: ( a + b i ) + ( c + d i ) = ( a + c ) + ( b + d ) i (a + bi) + (c + di) = (a + c) + (b + d)i ( a + bi ) + ( c + d i ) = ( a + c ) + ( b + d ) i and ( a + b i ) ( c + d i ) = ( a c − b d ) + ( a d + b c ) i (a + bi)(c + di) = (ac - bd) + (ad + bc)i ( a + bi ) ( c + d i ) = ( a c − b d ) + ( a d + b c ) i .
Proposition 1.1 (Properties of Complex Arithmetic). For all z , w ∈ C z, w \in \mathbb{C} z , w ∈ C :
z + w = w + z z + w = w + z z + w = w + z and z w = w z zw = wz z w = w z (commutativity)( z + w ) + u = z + ( w + u ) (z + w) + u = z + (w + u) ( z + w ) + u = z + ( w + u ) and ( z w ) u = z ( w u ) (zw)u = z(wu) ( z w ) u = z ( w u ) (associativity)z ( w + u ) = z w + z u z(w + u) = zw + zu z ( w + u ) = z w + z u (distributivity)There exist additive identity 0 0 0 and multiplicative identity 1 1 1 . Every z ≠ 0 z \neq 0 z = 0 has a multiplicative inverse 1 z = z ˉ ∣ z ∣ 2 \frac{1}{z} = \frac{\bar{z}}{|z|^2} z 1 = ∣ z ∣ 2 z ˉ . Remark. The complex field C \mathbb{C} C cannot be ordered: there is no total ordering on C \mathbb{C} C Compatible with the field operations. In particular, i 2 = − 1 i^2 = -1 i 2 = − 1 precludes any such ordering.
Definition. The complex conjugate of z = a + b i z = a + bi z = a + bi is z ˉ = a − b i \bar{z} = a - bi z ˉ = a − bi .
Proposition 1.2. For all z , w ∈ C z, w \in \mathbb{C} z , w ∈ C :
z + w ‾ = z ˉ + w ˉ \overline{z + w} = \bar{z} + \bar{w} z + w = z ˉ + w ˉ and z w ‾ = z ˉ w ˉ \overline{zw} = \bar{z}\bar{w} z w = z ˉ w ˉ z z ˉ = ∣ z ∣ 2 z\bar{z} = |z|^2 z z ˉ = ∣ z ∣ 2 z + z ˉ = 2 R e ( z ) z + \bar{z} = 2\,\mathrm{Re}(z) z + z ˉ = 2 Re ( z ) and z − z ˉ = 2 i I m ( z ) z - \bar{z} = 2i\,\mathrm{Im}(z) z − z ˉ = 2 i Im ( z ) z ˉ ˉ = z \bar{\bar{z}} = z z ˉ ˉ = z Definition. The modulus (or absolute value) of z = a + b i z = a + bi z = a + bi is ∣ z ∣ = a 2 + b 2 |z| = \sqrt{a^2 + b^2} ∣ z ∣ = a 2 + b 2 .
Proposition 1.3 (Modulus Properties). For all z , w ∈ C z, w \in \mathbb{C} z , w ∈ C :
∣ z ∣ ≥ 0 |z| \geq 0 ∣ z ∣ ≥ 0 with equality iff z = 0 z = 0 z = 0 ∣ z w ∣ = ∣ z ∣ ∣ w ∣ |zw| = |z||w| ∣ z w ∣ = ∣ z ∣∣ w ∣ ∣ z + w ∣ ≤ ∣ z ∣ + ∣ w ∣ |z + w| \leq |z| + |w| ∣ z + w ∣ ≤ ∣ z ∣ + ∣ w ∣ (triangle inequality)∣ ∣ z ∣ − ∣ w ∣ ∣ ≤ ∣ z − w ∣ \bigl||z| - |w|\bigr| \leq |z - w| ∣ z ∣ − ∣ w ∣ ≤ ∣ z − w ∣ (reverse triangle inequality)Proof of (3). ∣ z + w ∣ 2 = ( z + w ) ( z ˉ + w ˉ ) = ∣ z ∣ 2 + z w ˉ + z ˉ w + ∣ w ∣ 2 = ∣ z ∣ 2 + 2 R e ( z w ˉ ) + ∣ w ∣ 2 ≤ ∣ z ∣ 2 + 2 ∣ z ∣ ∣ w ∣ + ∣ w ∣ 2 = ( ∣ z ∣ + ∣ w ∣ ) 2 |z + w|^2 = (z + w)(\bar{z} + \bar{w}) = |z|^2 + z\bar{w} + \bar{z}w + |w|^2 = |z|^2 + 2\,\mathrm{Re}(z\bar{w}) + |w|^2 \leq |z|^2 + 2|z||w| + |w|^2 = (|z| + |w|)^2 ∣ z + w ∣ 2 = ( z + w ) ( z ˉ + w ˉ ) = ∣ z ∣ 2 + z w ˉ + z ˉ w + ∣ w ∣ 2 = ∣ z ∣ 2 + 2 Re ( z w ˉ ) + ∣ w ∣ 2 ≤ ∣ z ∣ 2 + 2∣ z ∣∣ w ∣ + ∣ w ∣ 2 = ( ∣ z ∣ + ∣ w ∣ ) 2 . The inequality follows from R e ( z w ˉ ) ≤ ∣ z w ˉ ∣ = ∣ z ∣ ∣ w ∣ \mathrm{Re}(z\bar{w}) \leq |z\bar{w}| = |z||w| Re ( z w ˉ ) ≤ ∣ z w ˉ ∣ = ∣ z ∣∣ w ∣ . ■ \blacksquare ■
Every non-zero complex number can be written in polar form :
z = r ( cos θ + i sin θ ) = r e i θ z = r(\cos\theta + i\sin\theta) = re^{i\theta} z = r ( cos θ + i sin θ ) = r e i θ
Where r = ∣ z ∣ = a 2 + b 2 r = |z| = \sqrt{a^2 + b^2} r = ∣ z ∣ = a 2 + b 2 is the modulus and θ = arg ( z ) \theta = \arg(z) θ = arg ( z ) is the argument .
Definition. The principal argument A r g ( z ) \mathrm{Arg}(z) Arg ( z ) is the unique θ ∈ ( − π , π ] \theta \in (-\pi, \pi] θ ∈ ( − π , π ] Such that z = ∣ z ∣ e i θ z = |z|e^{i\theta} z = ∣ z ∣ e i θ . The argument arg ( z ) \arg(z) arg ( z ) is multi-valued: arg ( z ) = A r g ( z ) + 2 π k \arg(z) = \mathrm{Arg}(z) + 2\pi k arg ( z ) = Arg ( z ) + 2 π k for k ∈ Z k \in \mathbb{Z} k ∈ Z .
Proposition 1.4. If z 1 = r 1 e i θ 1 z_1 = r_1 e^{i\theta_1} z 1 = r 1 e i θ 1 and z 2 = r 2 e i θ 2 z_2 = r_2 e^{i\theta_2} z 2 = r 2 e i θ 2 Then z 1 z 2 = r 1 r 2 e i ( θ 1 + θ 2 ) z_1 z_2 = r_1 r_2 e^{i(\theta_1 + \theta_2)} z 1 z 2 = r 1 r 2 e i ( θ 1 + θ 2 ) and z 1 / z 2 = ( r 1 / r 2 ) e i ( θ 1 − θ 2 ) z_1/z_2 = (r_1/r_2)\, e^{i(\theta_1 - \theta_2)} z 1 / z 2 = ( r 1 / r 2 ) e i ( θ 1 − θ 2 ) .
Solution Problem. Convert z = − 1 + 3 i z = -1 + \sqrt{3}\,i z = − 1 + 3 i to polar form and find all arguments.
∣ z ∣ = ( − 1 ) 2 + ( 3 ) 2 = 1 + 3 = 2 |z| = \sqrt{(-1)^2 + (\sqrt{3})^2} = \sqrt{1 + 3} = 2 ∣ z ∣ = ( − 1 ) 2 + ( 3 ) 2 = 1 + 3 = 2 .
R e ( z ) = − 1 < 0 \mathrm{Re}(z) = -1 \lt 0 Re ( z ) = − 1 < 0 and I m ( z ) = 3 > 0 \mathrm{Im}(z) = \sqrt{3} \gt 0 Im ( z ) = 3 > 0 So z z z is in the second quadrant.
θ = arctan ( 3 − 1 ) = 2 π 3 \theta = \arctan\!\left(\frac{\sqrt{3}}{-1}\right) = \frac{2\pi}{3} θ = arctan ( − 1 3 ) = 3 2 π (adjusting to second quadrant).
Polar form: z = 2 e 2 π i / 3 z = 2\,e^{2\pi i/3} z = 2 e 2 π i /3 .
All arguments: arg ( z ) = 2 π 3 + 2 π k \arg(z) = \frac{2\pi}{3} + 2\pi k arg ( z ) = 3 2 π + 2 π k for k ∈ Z k \in \mathbb{Z} k ∈ Z .
Problem. Convert z = 3 e − i π / 4 z = 3e^{-i\pi/4} z = 3 e − iπ /4 to rectangular form.
z = 3 ( cos ( − π 4 ) + i sin ( − π 4 ) ) = 3 ( 2 2 − i 2 2 ) = 3 2 2 − 3 2 2 i z = 3\left(\cos\!\left(-\frac{\pi}{4}\right) + i\sin\!\left(-\frac{\pi}{4}\right)\right) = 3\left(\frac{\sqrt{2}}{2} - i\,\frac{\sqrt{2}}{2}\right) = \frac{3\sqrt{2}}{2} - \frac{3\sqrt{2}}{2}\,i z = 3 ( cos ( − 4 π ) + i sin ( − 4 π ) ) = 3 ( 2 2 − i 2 2 ) = 2 3 2 − 2 3 2 i .
Problem. Express z = − 3 − 4 i z = -3 - 4i z = − 3 − 4 i in polar form.
∣ z ∣ = 9 + 16 = 5 |z| = \sqrt{9 + 16} = 5 ∣ z ∣ = 9 + 16 = 5 .
Both real and imaginary parts are negative, so z z z is in the third quadrant.
θ = arctan ( 4 / 3 ) + π = π + arctan ( 4 / 3 ) \theta = \arctan(4/3) + \pi = \pi + \arctan(4/3) θ = arctan ( 4/3 ) + π = π + arctan ( 4/3 ) .
z = 5 e i ( π + arctan ( 4 / 3 ) ) z = 5\,e^{i(\pi + \arctan(4/3))} z = 5 e i ( π + a r c t a n ( 4/3 )) .
Euler’s formula: e i θ = cos θ + i sin θ e^{i\theta} = \cos\theta + i\sin\theta e i θ = cos θ + i sin θ .
De Moivre’s theorem: ( e i θ ) n = e i n θ (e^{i\theta})^n = e^{in\theta} ( e i θ ) n = e in θ So
( cos θ + i sin θ ) n = cos ( n θ ) + i sin ( n θ ) (\cos\theta + i\sin\theta)^n = \cos(n\theta) + i\sin(n\theta) ( cos θ + i sin θ ) n = cos ( n θ ) + i sin ( n θ )
Proposition 1.5. De Moivre’s theorem holds for all integers n n n Including negative values.
Proof. For n ≥ 0 n \geq 0 n ≥ 0 It follows by induction from the multiplication law e i α e i β = e i ( α + β ) e^{i\alpha}e^{i\beta} = e^{i(\alpha + \beta)} e i α e i β = e i ( α + β ) . For n < 0 n \lt 0 n < 0 Write n = − m n = -m n = − m with m > 0 m \gt 0 m > 0 : ( cos θ + i sin θ ) n = 1 ( cos θ + i sin θ ) m = 1 cos ( m θ ) + i sin ( m θ ) = cos ( − m θ ) + i sin ( − m θ ) = cos ( n θ ) + i sin ( n θ ) (\cos\theta + i\sin\theta)^n = \frac{1}{(\cos\theta + i\sin\theta)^m} = \frac{1}{\cos(m\theta) + i\sin(m\theta)} = \cos(-m\theta) + i\sin(-m\theta) = \cos(n\theta) + i\sin(n\theta) ( cos θ + i sin θ ) n = ( c o s θ + i s i n θ ) m 1 = c o s ( m θ ) + i s i n ( m θ ) 1 = cos ( − m θ ) + i sin ( − m θ ) = cos ( n θ ) + i sin ( n θ ) . ■ \blacksquare ■
Example. Compute ( 1 + i ) 20 (1 + i)^{20} ( 1 + i ) 20 .
1 + i = 2 e i π / 4 1 + i = \sqrt{2}\,e^{i\pi/4} 1 + i = 2 e iπ /4 So ( 1 + i ) 20 = ( 2 ) 20 e 20 π i / 4 = 2 10 e 5 π i = 1024 e π i = − 1024 (1 + i)^{20} = (\sqrt{2})^{20}\, e^{20\pi i/4} = 2^{10}\, e^{5\pi i} = 1024\,e^{\pi i} = -1024 ( 1 + i ) 20 = ( 2 ) 20 e 20 π i /4 = 2 10 e 5 π i = 1024 e π i = − 1024 .
Solution Problem. Express cos ( 5 θ ) \cos(5\theta) cos ( 5 θ ) in terms of cos θ \cos\theta cos θ using de Moivre.
By de Moivre: cos ( 5 θ ) + i sin ( 5 θ ) = ( cos θ + i sin θ ) 5 \cos(5\theta) + i\sin(5\theta) = (\cos\theta + i\sin\theta)^5 cos ( 5 θ ) + i sin ( 5 θ ) = ( cos θ + i sin θ ) 5 .
Expanding the right side by the binomial theorem and equating real parts:
cos ( 5 θ ) = cos 5 θ − 10 cos 3 θ sin 2 θ + 5 cos θ sin 4 θ \cos(5\theta) = \cos^5\theta - 10\cos^3\theta\sin^2\theta + 5\cos\theta\sin^4\theta cos ( 5 θ ) = cos 5 θ − 10 cos 3 θ sin 2 θ + 5 cos θ sin 4 θ .
Using sin 2 θ = 1 − cos 2 θ \sin^2\theta = 1 - \cos^2\theta sin 2 θ = 1 − cos 2 θ :
cos ( 5 θ ) = cos 5 θ − 10 cos 3 θ ( 1 − cos 2 θ ) + 5 cos θ ( 1 − cos 2 θ ) 2 \cos(5\theta) = \cos^5\theta - 10\cos^3\theta(1 - \cos^2\theta) + 5\cos\theta(1 - \cos^2\theta)^2 cos ( 5 θ ) = cos 5 θ − 10 cos 3 θ ( 1 − cos 2 θ ) + 5 cos θ ( 1 − cos 2 θ ) 2 = cos 5 θ − 10 cos 3 θ + 10 cos 5 θ + 5 cos θ − 10 cos 3 θ + 5 cos 5 θ = \cos^5\theta - 10\cos^3\theta + 10\cos^5\theta + 5\cos\theta - 10\cos^3\theta + 5\cos^5\theta = cos 5 θ − 10 cos 3 θ + 10 cos 5 θ + 5 cos θ − 10 cos 3 θ + 5 cos 5 θ = 16 cos 5 θ − 20 cos 3 θ + 5 cos θ = 16\cos^5\theta - 20\cos^3\theta + 5\cos\theta = 16 cos 5 θ − 20 cos 3 θ + 5 cos θ .
Problem. Show that ∑ k = 0 n − 1 cos ( k θ ) = sin ( n θ / 2 ) sin ( θ / 2 ) cos ( ( n − 1 ) θ 2 ) \sum_{k=0}^{n-1} \cos(k\theta) = \frac{\sin(n\theta/2)}{\sin(\theta/2)}\cos\!\left(\frac{(n-1)\theta}{2}\right) ∑ k = 0 n − 1 cos ( k θ ) = s i n ( θ /2 ) s i n ( n θ /2 ) cos ( 2 ( n − 1 ) θ ) For θ ∉ 2 π Z \theta \notin 2\pi\mathbb{Z} θ ∈ / 2 π Z .
Consider S = ∑ k = 0 n − 1 e i k θ = 1 − e i n θ 1 − e i θ S = \sum_{k=0}^{n-1} e^{ik\theta} = \frac{1 - e^{in\theta}}{1 - e^{i\theta}} S = ∑ k = 0 n − 1 e ik θ = 1 − e i θ 1 − e in θ (geometric series with r = e i θ ≠ 1 r = e^{i\theta} \neq 1 r = e i θ = 1 ).
S = e i n θ / 2 ( e − i n θ / 2 − e i n θ / 2 ) e i θ / 2 ( e − i θ / 2 − e i θ / 2 ) = e i ( n − 1 ) θ / 2 ⋅ sin ( n θ / 2 ) sin ( θ / 2 ) S = \frac{e^{in\theta/2}(e^{-in\theta/2} - e^{in\theta/2})}{e^{i\theta/2}(e^{-i\theta/2} - e^{i\theta/2})} = e^{i(n-1)\theta/2} \cdot \frac{\sin(n\theta/2)}{\sin(\theta/2)} S = e i θ /2 ( e − i θ /2 − e i θ /2 ) e in θ /2 ( e − in θ /2 − e in θ /2 ) = e i ( n − 1 ) θ /2 ⋅ s i n ( θ /2 ) s i n ( n θ /2 ) .
Taking real parts gives the result.
Definition. An n n n -th root of w ∈ C w \in \mathbb{C} w ∈ C is a complex number z z z such that z n = w z^n = w z n = w .
Proposition 1.6. Every non-zero w ∈ C w \in \mathbb{C} w ∈ C has exactly n n n distinct n n n -th roots. If w = ρ e i ϕ w = \rho\, e^{i\phi} w = ρ e i ϕ Then
z k = ρ 1 / n e i ( ϕ + 2 π k ) / n , k = 0 , 1 , … , n − 1 z_k = \rho^{1/n}\, e^{i(\phi + 2\pi k)/n}, \quad k = 0, 1, \ldots, n - 1 z k = ρ 1/ n e i ( ϕ + 2 π k ) / n , k = 0 , 1 , … , n − 1
Where ρ 1 / n > 0 \rho^{1/n} \gt 0 ρ 1/ n > 0 is the positive real n n n -th root of ρ \rho ρ .
Proof. If z n = w z^n = w z n = w Write z = r e i θ z = r\,e^{i\theta} z = r e i θ . Then r n e i n θ = ρ e i ϕ r^n e^{in\theta} = \rho\, e^{i\phi} r n e in θ = ρ e i ϕ So r = ρ 1 / n r = \rho^{1/n} r = ρ 1/ n and n θ = ϕ + 2 π k n\theta = \phi + 2\pi k n θ = ϕ + 2 π k . For k = 0 , 1 , … , n − 1 k = 0, 1, \ldots, n-1 k = 0 , 1 , … , n − 1 these give distinct Values of θ \theta θ ; for k ≥ n k \geq n k ≥ n they repeat. ■ \blacksquare ■
Remark. The n n n -th roots of w w w lie equally spaced on a circle of radius ρ 1 / n \rho^{1/n} ρ 1/ n Forming a Regular n n n -gon.
The n n n -th roots of unity are the solutions of z n = 1 z^n = 1 z n = 1 :
z k = e 2 π i k / n , k = 0 , 1 , … , n − 1 z_k = e^{2\pi i k / n}, \quad k = 0, 1, \ldots, n - 1 z k = e 2 π ik / n , k = 0 , 1 , … , n − 1
They form a regular n n n -gon on the unit circle in the complex plane.
Proposition 1.7. If ω = e 2 π i / n \omega = e^{2\pi i/n} ω = e 2 π i / n is a primitive n n n -th root of unity, then ∑ k = 0 n − 1 ω k = 0 \sum_{k=0}^{n-1} \omega^k = 0 ∑ k = 0 n − 1 ω k = 0 and ∑ k = 0 n − 1 ω j k = 0 \sum_{k=0}^{n-1} \omega^{jk} = 0 ∑ k = 0 n − 1 ω j k = 0 for any j j j not divisible by n n n .
Proof. The sum ∑ k = 0 n − 1 ω k = 1 − ω n 1 − ω = 1 − 1 1 − ω = 0 \sum_{k=0}^{n-1} \omega^k = \frac{1 - \omega^n}{1 - \omega} = \frac{1 - 1}{1 - \omega} = 0 ∑ k = 0 n − 1 ω k = 1 − ω 1 − ω n = 1 − ω 1 − 1 = 0 Provided ω ≠ 1 \omega \neq 1 ω = 1 . For j j j not divisible by n n n , ω j \omega^j ω j is a non-trivial root of unity, So the same argument applies. ■ \blacksquare ■
Solution Problem. Find all cube roots of − 8 -8 − 8 .
− 8 = 8 e i π -8 = 8\,e^{i\pi} − 8 = 8 e iπ . The cube roots are: z k = 8 1 / 3 e i ( π + 2 π k ) / 3 = 2 e i ( π + 2 π k ) / 3 z_k = 8^{1/3}\, e^{i(\pi + 2\pi k)/3} = 2\, e^{i(\pi + 2\pi k)/3} z k = 8 1/3 e i ( π + 2 π k ) /3 = 2 e i ( π + 2 π k ) /3 for k = 0 , 1 , 2 k = 0, 1, 2 k = 0 , 1 , 2 .
z 0 = 2 e i π / 3 = 2 ( 1 2 + i 3 2 ) = 1 + i 3 z_0 = 2\,e^{i\pi/3} = 2\left(\frac{1}{2} + i\,\frac{\sqrt{3}}{2}\right) = 1 + i\sqrt{3} z 0 = 2 e iπ /3 = 2 ( 2 1 + i 2 3 ) = 1 + i 3 . z 1 = 2 e i π = − 2 z_1 = 2\,e^{i\pi} = -2 z 1 = 2 e iπ = − 2 . z 2 = 2 e i 5 π / 3 = 2 ( 1 2 − i 3 2 ) = 1 − i 3 z_2 = 2\,e^{i5\pi/3} = 2\left(\frac{1}{2} - i\,\frac{\sqrt{3}}{2}\right) = 1 - i\sqrt{3} z 2 = 2 e i 5 π /3 = 2 ( 2 1 − i 2 3 ) = 1 − i 3 .
Problem. Find all fourth roots of z = 16 i z = 16i z = 16 i .
16 i = 16 e i π / 2 16i = 16\,e^{i\pi/2} 16 i = 16 e iπ /2 . The fourth roots are: z k = 16 1 / 4 e i ( π / 2 + 2 π k ) / 4 = 2 e i ( π / 8 + π k / 2 ) z_k = 16^{1/4}\, e^{i(\pi/2 + 2\pi k)/4} = 2\, e^{i(\pi/8 + \pi k/2)} z k = 1 6 1/4 e i ( π /2 + 2 π k ) /4 = 2 e i ( π /8 + π k /2 ) for k = 0 , 1 , 2 , 3 k = 0, 1, 2, 3 k = 0 , 1 , 2 , 3 .
z 0 = 2 e i π / 8 z_0 = 2\,e^{i\pi/8} z 0 = 2 e iπ /8 , z 1 = 2 e i 5 π / 8 z_1 = 2\,e^{i5\pi/8} z 1 = 2 e i 5 π /8 , z 2 = 2 e i 9 π / 8 z_2 = 2\,e^{i9\pi/8} z 2 = 2 e i 9 π /8 , z 3 = 2 e i 13 π / 8 z_3 = 2\,e^{i13\pi/8} z 3 = 2 e i 13 π /8 .
Problem. Show that the n n n -th roots of any non-zero w w w are in geometric progression.
The roots are z k = ρ 1 / n e i ( ϕ + 2 π k ) / n = z 0 ⋅ ( e 2 π i / n ) k = z 0 ⋅ ω k z_k = \rho^{1/n}\, e^{i(\phi + 2\pi k)/n} = z_0 \cdot \left(e^{2\pi i/n}\right)^k = z_0 \cdot \omega^k z k = ρ 1/ n e i ( ϕ + 2 π k ) / n = z 0 ⋅ ( e 2 π i / n ) k = z 0 ⋅ ω k Where ω = e 2 π i / n \omega = e^{2\pi i/n} ω = e 2 π i / n is a primitive n n n -th root of unity. This is a geometric sequence With ratio ω \omega ω .