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1.5 Complex Numbers (HL)

Tags
Complex number
Modulus
Argument
Imaginary number
Conjugate
De Moivre's theorem
The number ii, where i2=1i^2=-1. (1.12)
Cartesian form z=a+biz = a + bi ; the terms real part, imaginary part, conjugate, modulus and argument.
The complex plane.
Modulus–argument (polar) form: z=r(cosθ+isinθ)=rcisθz = r(\cosθ + i\sinθ) = rcisθ (1.13)
Euler form: z=reiθ z = re^{iθ}
Sums, products and quotients in Cartesian, polar or Euler forms and their geometric interpretation.
Complex conjugate roots of quadratic and polynomial equations with real coefficients. (1.14)
De Moivre’s theorem and its extension to rational exponents.
Powers and roots of complex numbers.
Terminology
Definition
Imaginary number ii
This is defined as i=1i=\sqrt{-1} with the property of i2=1i^2=-1.
Complex number zz
Any number of the form z=a+biz=a+bi, where a,bR,zCa, b ∈ ℝ, z∈ℂ. zz is purely imaginary if a0a ≠ 0.  We cal; Re(z)=aRe(z)=a as the real part, and Im(z)=biIm(z )=bi as the imaginary part. For mathematical operations, treat ii like a variable.  To achieve real denominator for the form a+bic+di\frac{a+bi}{c+di}, we need to perform rationalization. Multiply the conjugate of the denominator to both the numerator and the denominator.
a+bic+di(cdicdi)=(a+bi)(cdi)c2+d2\frac{a+bi}{c+di}(\frac{c-di}{c-di})=\frac{(a+bi)(c-di)}{c^2+d^2}
Conjugates
If z=a+biz = a+bi, then z=z=abi\overline{z}=z^*=a-bi is the complex conjugate of zz.
Argand Diagram (HL)
The Argand diagram is composed of the real axis and the imaginary axis. The vector OP=(xy)\overrightarrow {OP}=\binom xy represents x+yix+yi. In the complex plane, the conjugate of zz is a reflection of zz in the real axis.
Modulus (HL)
The modulus of z=a+biz=a+bi is z=a2+b2|z|=\sqrt{a^2+b^2}, i.e. the magnitude of the vector (ab)\binom ab. 1. z=z|\overline {z}|=|z| 2. z2=zz|z|^2=z\overline z 3.z1z2=z1z2 |z_1z_2|=|z_1||z_2| 4. z1z2=z1z2|\frac{z_1}{z_2}|=\frac{|z_1|}{|z_2|} 5. z1z2...zn=z1z2...zn,zn=zn|z_1z_2...z_n|=|z_1||z_2|...|z_n|, |z^n| = |z|^n Given z1=OP1z_1=\overrightarrow{OP_1} and z2=OP2z_2=\overrightarrow{OP_2}, the distance between P1P_1 and P2P_2 is z1z2|z_1-z_2|.
Argument (HL)
The argument of zz, arg zz is the angle in the interval πθπ-π ≤ \theta ≤ π which is measured anti-clockwise between the positive real axis and OP\overrightarrow {OP}. Real numbers have argument of 0 or ππ. Purely imaginary numbers have argument of π2\frac π2 or π2-\frac π2.
Polar form (HL)
On an Argand diagram, we can represent zz in terms of trigonometry:  z=rcosθ+irsinθ=r(cosθ+isinθ)=rcisθ=zcisθz = r\cos\theta+ir\sin\theta=r(\cos\theta+i\sin\theta) = rcis\theta=|z|cis\theta. For z=zcisθz=|z|cis\theta, z |z| is the modulus while is the argument.  1. cisθcisΨ=cis(θ+Ψ)cis\theta cis\Psi=cis(\theta+\Psi) 2. cisθcisΨ=cis(θΨ)\frac{cis\theta}{cis\Psi}=cis(\theta-\Psi) 3. cis(θ+k2π)=cisθcis(\theta+k2\pi)=cis\theta 4. z=zcis(θ)\overline z=|z|cis(-\theta) 5. If z=zcisθz=|z|cis\theta and w=wcisΨw=|w|cis\Psi, then zw=zwcis(θ+Ψ)zw=|z||w|cis(\theta+\Psi).
Euler form (HL)
z=rcosθ+irsinθ=r(cosθ+isinθ)=reiθz=r\cos\theta+ir\sin\theta=r(\cos\theta+i\sin\theta)=re^{i\theta}.
De Moivre's theorem (HL)
(zcisθ)n=zn(cisnθ)(|z| cis\theta )^n=|z|^n(cis n\theta) Utilize this to solve the equation zn=cz^n=c, expressing cc as a polar form.  The solutions to this equation will form a regular nn-polygon in the Argand diagram. Proof Preposition: (zcisθ)n=zn(cisnθ)(|z| cis\theta )^n=|z|^n(cis n\theta) for nZ+ n∈Z^+ Base case: For n=1n=1, (zcisθ)1=zcisθ(|z|cis\theta)^1=|z|cis\theta  Hence, n=1n=1 is true. Induction: Suppose n=kn=k is true, then (zcisθ)k=zkciskθ(|z|cis\theta)^k=|z|^kcisk\theta. Let n=k+1n=k+1, then (zcisθ)k+1(|z|cis\theta)^{k+1} =zkciskθzcisθ|z|^kcisk\theta \cdot |z|cis\theta =zk+1cis(kθ+θ)=zk+1cis(k+1)θ|z|^{k+1}cis(k\theta+\theta)=|z|^{k+1}cis(k+1)\theta Conclusion: n=1n=1 is true n=k+1n=k+1 is true whenever n=kn=k is true, hence the preposition is true for all nZ+ n∈Z^+.
As you delve more into mathematics, the Euler form is much more commonly used than the Polar form.