1: Complex numbers: extensions
Extensions
Natural numbers, N={0,1,2,…}{\mathbb{N} = \{0,1,2,\ldots \}}
Counting. Addition, multiplication is well-defined.
Integers, Z={0,±1,±2,…}\mathbb{Z} = \{0, \pm 1, \pm 2,\ldots \}
Subtraction is well-defined.
Rational numbers, Q={ab:a,b∈Z,b≠0}\mathbb{Q} = \left \{ \frac{a}{b}: a,b \in \mathbb{Z}, b \neq 0 \right \}
Division is well-defined.
Real numbers, R{\mathbb{R}}
0 x
π,e,x{\pi, e, \sqrt{x}} where x≥0.{x\geq 0.}
Complex numbers, C{\mathbb{C}}
‘‘i=−1"{``i=\sqrt{-1}"}
z=a+bi{z=a+bi}, where a,b∈R{a,b\in\mathbb{R}}
Real and imaginary parts
z=x+yi{z=x+yi}
Re(z)=x,Im(z)=y{\textrm{Re}(z)=x, \textrm{Im}(z)=y}
Powers of i{i}
i2=−1i^2 = -1
Addition/subtraction
(1+2i)−(3−5i)=−2+7i(1 + 2 i) - (3 - 5i ) = -2+7i
Multiplication
(2+i)(3+4i)=6+8i+3i+4i2=2+11i\begin{aligned} &(2 + i)(3 + 4i) \\ &= 6 + 8i + 3i + 4i ^ 2 \\ &=2+11i \end{aligned}
Conjugation
If z=x+yi{z=x + y i}, the complex conjugate, z∗,{z^*,} is given by z∗=x−yi.{z^*=x-yi.}
z+z∗=2x=2Re(z)z+z^* = 2x = 2 \textrm{Re}(z)
z−z∗=2yi=2Im(z)iz-z^* = 2yi = 2 \textrm{Im}(z)i
zz∗=x2+y2=∣z∣2\boxed{zz^* = x^2+y^2 = |z|^2}
Division
2+i3−4i=2+i3−4i×3+4i3+4i=2+11i25\begin{aligned} \frac{2+i}{3-4i} &= \frac{2+i}{3-4i}\times\frac{3+4i}{3+4i} \\ &= \frac{2+11i}{25}\end{aligned}
Example 1: linear equation
Solve the equation (2+3i)z+8+2i=1−i.{(2+3i)z+8+2i=1-i.}
(2+3i)z=−7−3iz=−7−3i2+3i=−7−3i2+3i×2−3i2−3i=−23+15i13\begin{aligned} (2+3i)z &= -7-3i \\ z &= \frac{-7-3i}{2+3i} \\ &= \frac{-7-3i}{2+3i} \times \frac{2-3i}{2-3i} \\ &= \frac{-23+15i}{13} \end{aligned}
Example 2: quadratic equation
Solve the equation 2z2−2z+5=0.{2z^2-2z+5=0.}
z=−b±b2−4ac2a=2±4−404=2±−364=2±6i4=1±3i2\begin{aligned} z &= \frac{-b\pm\sqrt{b^2-4ac}}{2a} \\ &= \frac{2\pm \sqrt{4-40}}{4} \\ &= \frac{2\pm \sqrt{-36}}{4} \\ &= \frac{2\pm 6i}{4} \\ &= \frac{1\pm 3i}{2} \end{aligned}
Example 3: comparing real and imaginary parts
Solve the equation −18iz+zz∗=−56−90i.{-18iz+zz^*=-56-90i.}
Let z=x+yi{z=x+yi}
−18i(x+yi)+(x+yi)(x−yi)=−56−90i-18i{(x+yi)}+{(x+yi)}{(x-yi)}={-56-90i}
−18xi+18y+x2+y2=−56−90i-18xi+18y+x^2+y^2=-56-90i
By comparing imaginary parts: −18x=−90.{-18x=-90.}
Hence x=5.{x=5.}
By comparing real parts: 18y+x2+y2=−56.{18y+x^2+y^2=-56.}
y2+18y+81=0{y^2+18y+81=0}
(y+9)2=0{(y+9)^2=0}
y=−9{y=-9}
Hence z=5−9i{z=5-9i}
Factor theorem
If z=α{z=\alpha} is a root of a polynomial f(z){f(z)}, (i.e.f(α)=0){\left ( \mathrm{i.e. } f(\alpha)=0 \right )} then (z−α){(z-\alpha)} is a factor of f(z).{f(z).}
Fundamental theorem of algebra
The polynomial equation azn+bzn−1+…=0{az^n + bz^{n-1} + \ldots = 0} has n{n} complex roots (including multiplicity).
Conjugate root theorem
If z=a+bi{z=a+bi} is a root of the polynomial equation P(z)=0{P(z)=0} with real coefficients, then the complex conjugate z∗=a−bi{z^*=a-bi} is also a root.
Example 4: conjugate root theorem
It is given that 1+3i{1+3i} is a root of 2z3−z2+14z+30=0.{2z^3 - z^2 + 14z + 30=0.} Solve the equation.
Since all the coefficients are real, 1−3i{1-3i} is also a root.
(z−1+3i){(z-1+3i)}, (z−1−3i){(z-1-3i)} are factors of the cubic polynomial.
2z3−z2+14z+30=(z−1+3i)(z−1−3i)(az+b){2z^3 - z^2 + 14z + 30} = {(z-1+3i)}\allowbreak {(z-1-3i)} \allowbreak {(az+b)}
2z3−z2+14z+30=(z2−2z+10)(az+b){2z^3 - z^2 + 14z + 30} = {(z^2-2z+10)}\allowbreak {(az+b)}
By comparing coefficients (or by long division): a=2,b=3.{a=2,b=3.}
Hence (z−1−3i)(z−1+3i)(2z+3)=0(z-1-3i) \allowbreak (z-1+3i) \allowbreak (2z+3)=0
z=1+3i{{z=1+3i}},z=1−3i{{z=1-3i}} or z=−1.5.{z=-1.5.}
0 x y Re(z) Im(z)
0 x y Re(z) Im(z) r θ
Convert cartesian to polar form
r=x2+y2{r=\sqrt{x^2+y^2}}
tan⁡θ=yx{\tan \theta = \frac{y}{x}}
α=tan⁡−1∣yx∣{\alpha = \tan^{-1} \left | \frac{y}{x} \right |}
θ={αif x>0,y>0π−αif x<0,y>0−(π−α)if x<0,y<0−αif x>0,y<0\theta = \begin{cases} \alpha &\text{if } x > 0, y > 0 \\ \pi - \alpha &\text{if } x < 0, y > 0 \\ -(\pi - \alpha) &\text{if } x < 0, y < 0 \\ -\alpha &\text{if } x > 0, y < 0 \end{cases}
Convert polar to cartesian form
x=rcos⁡θ{x=r\cos \theta}
y=rsin⁡θ{y=r \sin \theta}
Cartesian form
z=x+yi{z=x+yi}
Polar/trigo form
z=r(cos⁡θ+isin⁡θ){z=r(\cos \theta + i \sin \theta)}
Polar/exp form
z=reiθ{z=re^{i \theta}}
Example 5: conversion
z=−1−ir=12+12θ=−(π−π4)z=2e−i3π4\begin{aligned} z &= -1-i \\ r &= \sqrt{1^2 + 1^2} \\ \theta &= {\textstyle -\left (\pi - \frac{\pi}{4} \right ) }\\ z &= \sqrt{2} e^{-i \frac{3\pi}{4}} \end{aligned}
z=2ei2π3=2(cos⁡2π3+isin⁡2π3)=−1+3i\begin{aligned} z &= 2 e^{i \frac{2\pi}{3}} \\ &= {\textstyle 2 \left (\cos \frac{2\pi}{3} + i \sin \frac{2\pi}{3} \right )} \\ &= -1 + \sqrt{3}i \end{aligned}
Exponential form
(r1eiθ1)(r2eiθ2)=r1r2ei(θ1+θ2)r1eiθ1r2eiθ2=r1r2ei(θ1−θ2)(reiθ)n=rnei(nθ)(reiθ)∗=rei(−θ)\begin{aligned} \left ( r_1 e^{i \theta_1} \right ) \left ( r_2 e^{i \theta_2} \right ) &= r_1r_2 e^{i (\theta_1+\theta_2)} \\ \frac{r_1 e^{i \theta_1}}{r_2 e^{i \theta_2}} &= \frac{r_1}{r_2} e^{i (\theta_1-\theta_2)} \\ \left ( r e^{i \theta}\right )^n &= r^n e^{i (n \theta)} \\ \left ( r e^{i \theta}\right )^* &= r e^{i (- \theta)} \end{aligned}
Modulus
∣wz∣=∣w∣∣z∣∣wz∣=∣w∣∣z∣∣zn∣=∣z∣n∣z∗∣=∣z∣\begin{aligned} \left | w z \right | &= \left | w \right | \left | z \right | \\ \left | \frac{w}{z} \right | &= \frac{\left | w \right |}{\left | z \right |} \\ \left | z^n \right | &= \left | z \right |^n \\ \left | z^* \right | &= \left | z \right | \end{aligned}
Argument
arg⁡(wz)=arg⁡(w)+arg⁡(z)arg⁡(wz)=arg⁡(w)−arg⁡(z)arg⁡(zn)=narg⁡(z)arg⁡(z∗)=−arg⁡(z)\begin{aligned} \arg \left ( w z \right ) &= \arg \left ( w \right ) + \arg \left ( z \right ) \\ \arg \left ( \frac{w}{z} \right ) &= \arg \left ( w \right ) - \arg \left ( z \right ) \\ \arg \left ( z^n \right ) &= n \arg \left ( z \right ) \\ \arg \left ( z^* \right ) &= - \arg \left ( z \right ) \end{aligned}
Principal values
We typically leave our arguments in the range −π<θ≤π.{-\pi < \theta \leq \pi.}
Values outside of this range can be converted by adding/subtracting 2kπ{2k\pi} where k∈Z.{k \in \mathbb{Z}.}
Example 6: polar form arithmetic
Evaluate ((e−π12i)7e−5π6ieπ3i)∗{\left ( \frac{ \left ( e^{-\frac{\pi}{12}i} \right )^7 e^{-\frac{5\pi}{6}i} }{ e^{\frac{\pi}{3}i} } \right )^*}
z=(e−7π12ie−5π6ieπ3i)∗=(e−17π12ieπ3i)∗=(e−21π12i)∗=e7π4i=e−π4i\begin{aligned} z &= \left ( \frac{ e^{-\frac{7\pi}{12}i} e^{-\frac{5\pi}{6}i} }{ e^{\frac{\pi}{3}i} } \right )^* = \left ( \frac{ e^{-\frac{17\pi}{12}i} }{ e^{\frac{\pi}{3}i} } \right )^* \\ &= \left ( e^{-\frac{21\pi}{12}i} \right )^* = e^{\frac{7\pi}{4}i} \\ &= e^{-\frac{\pi}{4}i} \end{aligned}
Condition Cartesian form, x+yi{x+yi} Polar form, reiθ{re^{i\theta}}
Real y=0{y=0} θ=kπ,k∈Z{\theta = k \pi,} \allowbreak {k \in \mathbb{Z}}
Real and positive y=0,x>0y=0, \allowbreak x > 0 θ=2kπ,k∈Z{\theta = 2k \pi,} \allowbreak k \in \mathbb{Z}
Real and negative y=0,x<0y=0, \allowbreak x < 0 θ=π+2kπ,k∈Z{\theta = \pi + 2k \pi,} \allowbreak k \in \mathbb{Z}
Purely imaginary x=0{x = 0} θ=π2+kπ,k∈Z{\theta = \frac{\pi}{2} + k \pi,} \allowbreak k \in \mathbb{Z}
0 Re(z) Im(z)
Example 7: purely imaginary
Find the three smallest positive integers n{n} such that (e−π12i)n{\left ( e^{-\frac{\pi}{12}i} \right )^n} is purely imaginary
−nπ12=π2+kπ{-\frac{n\pi}{12} = \frac{\pi}{2} + k \pi}
n=−6−12k{n = -6-12k}
Smallest n=6,18,24{n = 6, 18, 24} (when k=−1,−2,−3.{k=-1,-2,-3.})
Example 8: exam-style question
The complex number z{z} is given by z=3+bi{z=3+bi}, where b{b} is a real number.
(a) Find the possible values of b{b} if z2z∗{\displaystyle \frac{z^2}{z^*}} is real.
For the rest of the question, it is further given that b>0{b>0}.
(b) Find the smallest integer value of n{n} such that ∣zn∣>1000{\left | z^n \right | > 1000}.
(c) For the value of n{n} found in (b), find the values of ∣zn∣{\left | z^n \right |} and arg⁡(zn){\arg\left(z^n\right)}, where −π<aarg⁡(zn)≤π{-\pi <a \arg \left ( z^n \right) \leq \pi}.
(d) On a single Argand diagram mark out the points A,B,C,D{A,B,C,D} and E{E} representing the complex numbers z{z}, z2z∗{\frac{z^2}{z^*}}, z∗{z^*}, 18z{\frac{18}{z}} and z26{\frac{z^2}{6}} respectively.
We first evaluate z2z∗{\frac{z^2}{z^*}} in terms of b{b}
z2z∗=(3+bi)23−bi=9−b2+6bi3−bi⋅3+bi3+bi=3(9−b2)−6b2+(18b+(9−b2)b)i9+b2\begin{aligned} \frac{z^2}{z^*} &= \frac{\left (3+bi \right)^2}{3-bi} \\ &= \frac{9-b^2+6bi}{3-bi} \cdot \frac{3+bi}{3+bi} \\ &= \frac{3(9-b^2)-6b^2+\left( 18b+(9-b^2)b \right)i}{9+b^2} \end{aligned}
Since z2z∗{\frac{z^2}{z^*}} is real,
18b+(9−b2)b=0{18b+(9-b^2)b=0}
b(27−b2)=0{b(27-b^2)=0}
b=0,b=±33{b=0, b=\pm 3\sqrt{3}}
b=33{b=3\sqrt{3}} since b>0{b>0}
∣zn∣>1000∣3+33i∣n>1000(32+(33)2)n>1000nln⁡6>ln⁡1000n>3.8553\begin{aligned} \left | z^n \right | &> 1000 \\ \left | 3 + 3\sqrt{3}i \right |^n &> 1000 \\ \left ( \sqrt{3^2 + (3\sqrt{3})^2} \right )^n &> 1000 \\ n \ln 6 &> \ln 1000 \\ n &> 3.8553 \end{aligned}
Smallest integer value of n=4{n=4}
∣zn∣=64=1296{\left |z^n \right | = 6^4 = 1296}
arg⁡(z)=tan⁡−1(333)=π3{\arg \left ( z \right ) = \tan^{-1} \left ( \frac{3\sqrt{3}}{3} \right) = \frac{\pi}{3}}
arg⁡(zn)=4arg⁡(z)≡4π3−2π=−2π3{\arg \left ( z^n \right ) = 4 \arg (z) \equiv \frac{4\pi}{3}-2\pi = -\frac{2\pi}{3} }
z=6eiπ3,z2z∗=36ei2π36e−iπ3=6eiπ,z∗=6e−iπ3,{z=6e^{i\frac{\pi}{3}}, \quad \frac{z^2}{z^*} = \frac{36e^{i\frac{2\pi}{3}}}{6e^{-i\frac{\pi}{3}}} = 6e^{i\pi}, \quad z^* = 6e^{-i\frac{\pi}{3}}, }
18z=186eiπ3=3e−iπ3,z26=36ei2π36=6ei2π3{\frac{18}{z} = \frac{18}{6e^{i\frac{\pi}{3}}} = 3e^{-i\frac{\pi}{3}}, \quad \frac{z^2}{6} = \frac{36e^{i\frac{2\pi}{3}}}{6} = 6e^{i\frac{2\pi}{3}} }
A C E D Re(z) B Im(z) 6 6 3 π / 3 π / 3 π / 3
Conjugate formulas
z+z∗=2x=2rcos⁡θ{z+z^* = 2x = 2r \cos \theta}
z−z∗=2yi=2risin⁡θ{z-z^* = 2yi = 2r i \sin \theta}
The half-angle 'trick'
eiθ+1=eiθ2eiθ2+eiθ2e−iθ2=eiθ2(eiθ2+e−iθ2)=eiθ2(2cos⁡θ2)=(2cos⁡θ2)eiθ2\begin{aligned} e^{i \theta} + 1 &= e^{i \frac{\theta}{2}}e^{i \frac{\theta}{2}} + e^{i \frac{\theta}{2}}e^{-i \frac{\theta}{2}} \\ &= e^{i \frac{\theta}{2}} \left ( e^{i \frac{\theta}{2}} + e^{-i \frac{\theta}{2}} \right) \\ &= e^{i \frac{\theta}{2}} \left ( 2 \cos \frac{\theta}{2} \right) \\ &= \left ( 2 \cos \frac{\theta}{2} \right) e^{i \frac{\theta}{2}} \end{aligned}
eiθ−1=eiθ2eiθ2−eiθ2e−iθ2=eiθ2(eiθ2−e−iθ2)=eiθ2(2isin⁡θ2)=eiθ2(2eiπ2sin⁡θ2)=(2sin⁡θ2)ei(π2+θ2)\begin{aligned} e^{i \theta} - 1 &= e^{i \frac{\theta}{2}}e^{i \frac{\theta}{2}} - e^{i \frac{\theta}{2}}e^{-i \frac{\theta}{2}} \\ &= e^{i \frac{\theta}{2}} \left ( e^{i \frac{\theta}{2}} - e^{-i \frac{\theta}{2}} \right) \\ &= e^{i \frac{\theta}{2}} \left ( 2 i \sin \frac{\theta}{2} \right) \\ &= e^{i \frac{\theta}{2}} \left ( 2 e^{i \frac{\pi}{2}} \sin \frac{\theta}{2} \right) \\ &= \left ( 2 \sin \frac{\theta}{2} \right) e^{i \left ( \frac{\pi}{2} + \frac{\theta}{2} \right ) } \end{aligned}
Notes for printing (pdf):