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Complex Numbers

Complex AnalysisFoundations🟒 Free Lesson

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Why It Matters

Why It Matters

Complex numbers are far more than an abstract extension of the real number system. They are the natural language for describing oscillations, rotations, and wave phenomena. Every alternating current in an electrical circuit, every quantum mechanical wavefunction, every signal transmitted through fiber optics relies on complex arithmetic. The fundamental theorem of algebra β€” that every polynomial of degree n has exactly n roots in the complex plane β€” guarantees that complex numbers are algebraically complete. Without them, Fourier analysis, control theory, fluid dynamics, and much of modern physics would be impossible. Complex numbers unify exponential and trigonometric functions through Euler's formula, transforming difficult calculus problems into elegant algebra. Mastering complex numbers is the essential first step toward contour integration, residue calculus, and the powerful applications that follow.


Core Definitions

DfComplex Number

A complex number is an ordered pair (a,b)(a, b) of real numbers, written z=a+biz = a + bi, where ii is the imaginary unit satisfying i2=βˆ’1i^2 = -1. The real part is Re⁑(z)=a\operatorname{Re}(z) = a and the imaginary part is Im⁑(z)=b\operatorname{Im}(z) = b. The set of all complex numbers is denoted C\mathbb{C}.

DfConjugate

The complex conjugate of z=a+biz = a + bi is zβ€Ύ=aβˆ’bi\overline{z} = a - bi. The conjugate reflects zz across the real axis in the complex plane.

DfModulus (Absolute Value)

The modulus of z=a+biz = a + bi is ∣z∣=a2+b2|z| = \sqrt{a^2 + b^2}, representing the distance from the origin to zz in the complex plane.

DfArgument

The argument of z=a+biz = a + bi is the angle ΞΈ\theta such that z=∣z∣(cos⁑θ+isin⁑θ)z = |z|(\cos\theta + i\sin\theta). The principal argument Arg⁑(z)∈(βˆ’Ο€,Ο€]\operatorname{Arg}(z) \in (-\pi, \pi] is the unique value in this range. The general argument is arg⁑(z)=Arg⁑(z)+2kΟ€\arg(z) = \operatorname{Arg}(z) + 2k\pi for any integer kk.

DfPolar Form

Any nonzero complex number can be written as z=r(cos⁑θ+isin⁑θ)=reiθz = r(\cos\theta + i\sin\theta) = re^{i\theta}, where r=∣z∣r = |z| and θ=arg⁑(z)\theta = \arg(z). This representation is called the polar form or exponential form.

DfRoots of Unity

The nn-th roots of unity are the solutions to zn=1z^n = 1. They are given by zk=e2Ο€ik/nz_k = e^{2\pi ik/n} for k=0,1,2,…,nβˆ’1k = 0, 1, 2, \ldots, n-1. These are equally spaced on the unit circle at angles 2Ο€k/n2\pi k/n.


Key Formulas

Complex Number Arithmetic

z1+z2=(a+c)+(b+d)i,z1β‹…z2=(acβˆ’bd)+(ad+bc)iz_1 + z_2 = (a+c) + (b+d)i, \quad z_1 \cdot z_2 = (ac - bd) + (ad + bc)i

Here,

  • z1=a+biz_1 = a + bi=First complex number
  • z2=c+diz_2 = c + di=Second complex number

Modulus

∣z∣=a2+b2,∣z∣2=zβ‹…zβ€Ύ|z| = \sqrt{a^2 + b^2}, \quad |z|^2 = z \cdot \overline{z}

Here,

  • z=a+biz = a + bi=Complex number
  • ∣z∣|z|=Distance from origin

Division of Complex Numbers

z1z2=(a+bi)(cβˆ’di)c2+d2=ac+bdc2+d2+bcβˆ’adc2+d2i\frac{z_1}{z_2} = \frac{(a+bi)(c-di)}{c^2 + d^2} = \frac{ac+bd}{c^2+d^2} + \frac{bc-ad}{c^2+d^2}i

Here,

  • z1=a+biz_1 = a+bi=Numerator
  • z2=c+diz_2 = c+di=Denominator (nonzero)

Polar Form

z=reiθ,r=∣z∣,θ=arg⁑(z)z = re^{i\theta}, \quad r = |z|, \quad \theta = \arg(z)

Here,

  • rr=Modulus (magnitude) of z
  • ΞΈ\theta=Argument (angle) of z in radians

Euler's Formula

eiθ=cos⁑θ+isin⁑θe^{i\theta} = \cos\theta + i\sin\theta

Here,

  • ΞΈ\theta=Angle in radians
  • eiΞΈe^{i\theta}=Point on unit circle at angle \theta

De Moivre's Theorem

(reiθ)n=rneinθ=rn(cos⁑nθ+isin⁑nθ)(re^{i\theta})^n = r^n e^{in\theta} = r^n(\cos n\theta + i\sin n\theta)

Here,

  • rr=Modulus of original complex number
  • ΞΈ\theta=Argument of original complex number
  • nn=Integer exponent

N-th Roots

z1/n=r1/nei(ΞΈ+2kΟ€)/n,k=0,1,…,nβˆ’1z^{1/n} = r^{1/n} e^{i(\theta + 2k\pi)/n}, \quad k = 0, 1, \ldots, n-1

Here,

  • r1/nr^{1/n}=Real n-th root of the modulus
  • ΞΈ+2kΟ€\theta + 2k\pi=Argument offset by full rotations

Trigonometric Identities via Euler

cos⁑θ=eiΞΈ+eβˆ’iΞΈ2,sin⁑θ=eiΞΈβˆ’eβˆ’iΞΈ2i\cos\theta = \frac{e^{i\theta} + e^{-i\theta}}{2}, \quad \sin\theta = \frac{e^{i\theta} - e^{-i\theta}}{2i}

Here,

  • eiΞΈe^{i\theta}=Exponential form of rotation
  • eβˆ’iΞΈe^{-i\theta}=Conjugate rotation

Polar Multiplication

z1z2=r1r2ei(ΞΈ1+ΞΈ2)z_1 z_2 = r_1 r_2 e^{i(\theta_1 + \theta_2)}

Here,

  • r1,r2r_1, r_2=Moduli of the two numbers
  • ΞΈ1,ΞΈ2\theta_1, \theta_2=Arguments of the two numbers

Important Theorems

ThEuler's Formula

For any real number θ\theta, eiθ=cos⁑θ+isin⁑θe^{i\theta} = \cos\theta + i\sin\theta.

Proof Sketch: Start with the Taylor series for exe^x, cos⁑x\cos x, and sin⁑x\sin x. Substitute x=iΞΈx = i\theta into the exponential series. Since i2=βˆ’1i^2 = -1, i3=βˆ’ii^3 = -i, i4=1i^4 = 1, the series splits into real (cosine) and imaginary (sine) parts, yielding the result. β–‘\square

Key Consequence: eiΟ€+1=0e^{i\pi} + 1 = 0 (Euler's identity), connecting the five most important constants in mathematics.

ThFundamental Theorem of Algebra

Every non-constant polynomial p(z)=anzn+anβˆ’1znβˆ’1+β‹―+a0p(z) = a_n z^n + a_{n-1} z^{n-1} + \cdots + a_0 with complex coefficients has at least one root in C\mathbb{C}. Equivalently, p(z)p(z) factors completely as p(z)=an(zβˆ’z1)(zβˆ’z2)β‹―(zβˆ’zn)p(z) = a_n(z - z_1)(z - z_2) \cdots (z - z_n) where z1,…,znz_1, \ldots, z_n are the roots (counted with multiplicity).

Implication: C\mathbb{C} is algebraically closed β€” no polynomial equation lacks solutions. This is the essential reason complex numbers are so powerful in analysis.

ThDe Moivre's Theorem

For any real number r>0r > 0, any angle ΞΈ\theta, and any integer nn:

(reiΞΈ)n=rneinΞΈ(re^{i\theta})^n = r^n e^{in\theta}

Proof Sketch: Use induction on nn. For n=1n = 1, trivial. Assume true for nn. Then (reiΞΈ)n+1=(reiΞΈ)nβ‹…reiΞΈ=rneinΞΈβ‹…reiΞΈ=rn+1ei(n+1)ΞΈ(re^{i\theta})^{n+1} = (re^{i\theta})^n \cdot re^{i\theta} = r^n e^{in\theta} \cdot re^{i\theta} = r^{n+1} e^{i(n+1)\theta}. β–‘\square

ThProperties of Modulus and Argument

For z1,z2∈Cz_1, z_2 \in \mathbb{C}:

  1. ∣z1z2∣=∣z1βˆ£β‹…βˆ£z2∣|z_1 z_2| = |z_1| \cdot |z_2|
  2. ∣z1+z2βˆ£β‰€βˆ£z1∣+∣z2∣|z_1 + z_2| \leq |z_1| + |z_2| (Triangle Inequality)
  3. arg⁑(z1z2)=arg⁑(z1)+arg⁑(z2)(mod2Ο€)\arg(z_1 z_2) = \arg(z_1) + \arg(z_2) \pmod{2\pi}
  4. arg⁑(z1/z2)=arg⁑(z1)βˆ’arg⁑(z2)(mod2Ο€)\arg(z_1 / z_2) = \arg(z_1) - \arg(z_2) \pmod{2\pi}
  5. zβ‹…zβ€Ύ=∣z∣2z \cdot \overline{z} = |z|^2

ThRoots of Unity Form a Group

The nn-th roots of unity {1,Ο‰,Ο‰2,…,Ο‰nβˆ’1}\{1, \omega, \omega^2, \ldots, \omega^{n-1}\} where Ο‰=e2Ο€i/n\omega = e^{2\pi i/n} form a cyclic group under multiplication. They are vertices of a regular nn-gon inscribed in the unit circle. Any polynomial znβˆ’1z^n - 1 factors as znβˆ’1=∏k=0nβˆ’1(zβˆ’Ο‰k)z^n - 1 = \prod_{k=0}^{n-1}(z - \omega^k).


Worked Examples

Example 1: Basic Complex Arithmetic

Let z1=3+4iz_1 = 3 + 4i and z2=1βˆ’2iz_2 = 1 - 2i.

Addition: z1+z2=(3+1)+(4βˆ’2)i=4+2iz_1 + z_2 = (3+1) + (4-2)i = 4 + 2i

Multiplication: z1z2=(3)(1)βˆ’(4)(βˆ’2)+[(3)(βˆ’2)+(4)(1)]i=3+8+(βˆ’6+4)i=11βˆ’2iz_1 z_2 = (3)(1) - (4)(-2) + [(3)(-2) + (4)(1)]i = 3 + 8 + (-6 + 4)i = 11 - 2i

Division: z1z2=(3+4i)(1+2i)(1βˆ’2i)(1+2i)=3+6i+4i+8i21+4=3βˆ’8+10i5=βˆ’5+10i5=βˆ’1+2i\frac{z_1}{z_2} = \frac{(3+4i)(1+2i)}{(1-2i)(1+2i)} = \frac{3 + 6i + 4i + 8i^2}{1 + 4} = \frac{3 - 8 + 10i}{5} = \frac{-5 + 10i}{5} = -1 + 2i

Modulus: ∣z1∣=32+42=25=5|z_1| = \sqrt{3^2 + 4^2} = \sqrt{25} = 5

Conjugate: z2β€Ύ=1+2i\overline{z_2} = 1 + 2i

Verification: z2β‹…z2β€Ύ=(1βˆ’2i)(1+2i)=1+4=5=∣z2∣2z_2 \cdot \overline{z_2} = (1-2i)(1+2i) = 1 + 4 = 5 = |z_2|^2 βœ“

Example 2: Converting to Polar Form

Convert z=βˆ’3+iz = -\sqrt{3} + i to polar form.

Step 1: r=∣z∣=(3)2+12=3+1=2r = |z| = \sqrt{(\sqrt{3})^2 + 1^2} = \sqrt{3 + 1} = 2

Step 2: tan⁑θ=1βˆ’3=βˆ’13\tan\theta = \frac{1}{-\sqrt{3}} = -\frac{1}{\sqrt{3}}. Since zz is in the second quadrant (Re⁑(z)<0\operatorname{Re}(z) < 0, Im⁑(z)>0\operatorname{Im}(z) > 0):

ΞΈ=Ο€βˆ’Ο€6=5Ο€6\theta = \pi - \frac{\pi}{6} = \frac{5\pi}{6}

Result: z=2eiβ‹…5Ο€/6=2(cos⁑5Ο€6+isin⁑5Ο€6)z = 2e^{i \cdot 5\pi/6} = 2\left(\cos\frac{5\pi}{6} + i\sin\frac{5\pi}{6}\right)

Verification: 2cos⁑(5Ο€/6)=2β‹…(βˆ’3/2)=βˆ’32\cos(5\pi/6) = 2 \cdot (-\sqrt{3}/2) = -\sqrt{3} βœ“ and 2sin⁑(5Ο€/6)=2β‹…(1/2)=12\sin(5\pi/6) = 2 \cdot (1/2) = 1 βœ“

Example 3: Computing Powers with De Moivre's Theorem

Compute (1+i)8(1 + i)^8.

Step 1: Convert to polar form. r=12+12=2r = \sqrt{1^2 + 1^2} = \sqrt{2}, ΞΈ=Ο€/4\theta = \pi/4 (first quadrant, tan⁑θ=1\tan\theta = 1).

So 1+i=2β‹…eiΟ€/41 + i = \sqrt{2} \cdot e^{i\pi/4}.

Step 2: Apply De Moivre's Theorem:

(1+i)8=(2)8β‹…eiβ‹…8Ο€/4=24β‹…eiβ‹…2Ο€=16β‹…eiβ‹…2Ο€=16(cos⁑2Ο€+isin⁑2Ο€)=16(1+i)^8 = (\sqrt{2})^8 \cdot e^{i \cdot 8\pi/4} = 2^4 \cdot e^{i \cdot 2\pi} = 16 \cdot e^{i \cdot 2\pi} = 16(\cos 2\pi + i\sin 2\pi) = 16

Alternative (brute force): (1+i)2=2i(1+i)^2 = 2i, (2i)2=βˆ’4(2i)^2 = -4, (βˆ’4)2=16(-4)^2 = 16. Same result. βœ“

Example 4: Finding N-th Roots of a Complex Number

Find all cube roots of z=8iz = 8i.

Step 1: Write zz in polar form. z=8i=8eiΟ€/2z = 8i = 8e^{i\pi/2} (modulus 8, argument Ο€/2\pi/2).

Step 2: Apply the nn-th root formula with n=3n = 3: z1/3=81/3β‹…ei(Ο€/2+2kΟ€)/3=2β‹…ei(Ο€/6+2kΟ€/3)z^{1/3} = 8^{1/3} \cdot e^{i(\pi/2 + 2k\pi)/3} = 2 \cdot e^{i(\pi/6 + 2k\pi/3)} for k=0,1,2k = 0, 1, 2.

Step 3: Compute each root:

  • k=0k = 0: w0=2eiΟ€/6=2(32+i2)=3+iw_0 = 2e^{i\pi/6} = 2\left(\frac{\sqrt{3}}{2} + \frac{i}{2}\right) = \sqrt{3} + i
  • k=1k = 1: w1=2ei(Ο€/6+2Ο€/3)=2ei5Ο€/6=2(βˆ’32+i2)=βˆ’3+iw_1 = 2e^{i(\pi/6 + 2\pi/3)} = 2e^{i5\pi/6} = 2\left(-\frac{\sqrt{3}}{2} + \frac{i}{2}\right) = -\sqrt{3} + i
  • k=2k = 2: w2=2ei(Ο€/6+4Ο€/3)=2ei3Ο€/2=2(0βˆ’i)=βˆ’2iw_2 = 2e^{i(\pi/6 + 4\pi/3)} = 2e^{i3\pi/2} = 2(0 - i) = -2i

Check: w03=(3+i)3=8iw_0^3 = (\sqrt{3}+i)^3 = 8i βœ“, w13=(βˆ’3+i)3=8iw_1^3 = (-\sqrt{3}+i)^3 = 8i βœ“, w23=(βˆ’2i)3=8iw_2^3 = (-2i)^3 = 8i βœ“

These three roots form an equilateral triangle inscribed in a circle of radius 2.

Example 5: Euler's Formula and Trigonometric Identities

Prove that cos⁑(3ΞΈ)=4cos⁑3ΞΈβˆ’3cos⁑θ\cos(3\theta) = 4\cos^3\theta - 3\cos\theta.

Step 1: By De Moivre's Theorem, (cos⁑θ+isin⁑θ)3=cos⁑3θ+isin⁑3θ(\cos\theta + i\sin\theta)^3 = \cos 3\theta + i\sin 3\theta.

Step 2: Expand the left side using the binomial theorem:

(cos⁑θ+isin⁑θ)3=cos⁑3ΞΈ+3icos⁑2ΞΈsin⁑θ+3i2cos⁑θsin⁑2ΞΈ+i3sin⁑3ΞΈ(\cos\theta + i\sin\theta)^3 = \cos^3\theta + 3i\cos^2\theta\sin\theta + 3i^2\cos\theta\sin^2\theta + i^3\sin^3\theta=cos⁑3ΞΈ+3icos⁑2ΞΈsinβ‘ΞΈβˆ’3cos⁑θsin⁑2ΞΈβˆ’isin⁑3ΞΈ= \cos^3\theta + 3i\cos^2\theta\sin\theta - 3\cos\theta\sin^2\theta - i\sin^3\theta=(cos⁑3ΞΈβˆ’3cos⁑θsin⁑2ΞΈ)+i(3cos⁑2ΞΈsinβ‘ΞΈβˆ’sin⁑3ΞΈ)= (\cos^3\theta - 3\cos\theta\sin^2\theta) + i(3\cos^2\theta\sin\theta - \sin^3\theta)

Step 3: Equate real parts: cos⁑3ΞΈ=cos⁑3ΞΈβˆ’3cos⁑θsin⁑2ΞΈ=cos⁑3ΞΈβˆ’3cos⁑θ(1βˆ’cos⁑2ΞΈ)=4cos⁑3ΞΈβˆ’3cos⁑θ\cos 3\theta = \cos^3\theta - 3\cos\theta\sin^2\theta = \cos^3\theta - 3\cos\theta(1 - \cos^2\theta) = 4\cos^3\theta - 3\cos\theta βœ“

Example 6: Roots of Unity and Polynomial Factorization

Factor z4βˆ’1z^4 - 1 over C\mathbb{C}.

Step 1: The 4th roots of unity are zk=e2Ο€ik/4z_k = e^{2\pi ik/4} for k=0,1,2,3k = 0, 1, 2, 3:

  • z0=1z_0 = 1
  • z1=eiΟ€/2=iz_1 = e^{i\pi/2} = i
  • z2=eiΟ€=βˆ’1z_2 = e^{i\pi} = -1
  • z3=ei3Ο€/2=βˆ’iz_3 = e^{i3\pi/2} = -i

Step 2: Factor:

z4βˆ’1=(zβˆ’1)(zβˆ’i)(z+1)(z+i)z^4 - 1 = (z - 1)(z - i)(z + 1)(z + i)

Step 3: Group into real quadratics: (zβˆ’1)(z+1)=z2βˆ’1(z-1)(z+1) = z^2 - 1 and (zβˆ’i)(z+i)=z2+1(z-i)(z+i) = z^2 + 1

So z4βˆ’1=(z2βˆ’1)(z2+1)=(zβˆ’1)(z+1)(z2+1)z^4 - 1 = (z^2 - 1)(z^2 + 1) = (z-1)(z+1)(z^2+1).

This demonstrates how complex roots enable complete factorization of polynomials.


Practice Problems

Problem 1: Complex Division

Compute (2+3i)(1βˆ’i)3+4i\frac{(2 + 3i)(1 - i)}{3 + 4i}.

Solution

Step 1: Numerator: (2+3i)(1βˆ’i)=2βˆ’2i+3iβˆ’3i2=2+i+3=5+i(2+3i)(1-i) = 2 - 2i + 3i - 3i^2 = 2 + i + 3 = 5 + i

Step 2: Divide by 3+4i3 + 4i by multiplying by conjugate:

5+i3+4i=(5+i)(3βˆ’4i)(3+4i)(3βˆ’4i)=15βˆ’20i+3iβˆ’4i29+16=15+4βˆ’17i25=19βˆ’17i25\frac{5+i}{3+4i} = \frac{(5+i)(3-4i)}{(3+4i)(3-4i)} = \frac{15 - 20i + 3i - 4i^2}{9 + 16} = \frac{15 + 4 - 17i}{25} = \frac{19 - 17i}{25}

Result: 1925βˆ’1725i\frac{19}{25} - \frac{17}{25}i

Problem 2: Polar Form Conversion

Express z=βˆ’1βˆ’i3z = -1 - i\sqrt{3} in polar form and find z6z^6.

Solution

Step 1: r=1+3=2r = \sqrt{1 + 3} = 2

Step 2: tan⁑θ=βˆ’3βˆ’1=3\tan\theta = \frac{-\sqrt{3}}{-1} = \sqrt{3}. Third quadrant: ΞΈ=Ο€+Ο€3=4Ο€3\theta = \pi + \frac{\pi}{3} = \frac{4\pi}{3}

Polar form: z=2ei4Ο€/3z = 2e^{i4\pi/3}

Step 3: z6=26β‹…eiβ‹…6β‹…4Ο€/3=64β‹…eiβ‹…8Ο€=64β‹…eiβ‹…0=64z^6 = 2^6 \cdot e^{i \cdot 6 \cdot 4\pi/3} = 64 \cdot e^{i \cdot 8\pi} = 64 \cdot e^{i \cdot 0} = 64

(Since 8Ο€8\pi is a multiple of 2Ο€2\pi, the exponential equals 1.)

Problem 3: Finding All Fifth Roots of Unity

Find all 5th roots of unity and express them in both exponential and rectangular form.

Solution

The 5th roots of unity are zk=e2Ο€ik/5z_k = e^{2\pi ik/5} for k=0,1,2,3,4k = 0, 1, 2, 3, 4.

Let Ο‰=e2Ο€i/5=cos⁑(2Ο€/5)+isin⁑(2Ο€/5)\omega = e^{2\pi i/5} = \cos(2\pi/5) + i\sin(2\pi/5). Using the exact values cos⁑(2Ο€/5)=5βˆ’14\cos(2\pi/5) = \frac{\sqrt{5}-1}{4} and sin⁑(2Ο€/5)=10+254\sin(2\pi/5) = \frac{\sqrt{10+2\sqrt{5}}}{4}:

  • z0=1z_0 = 1
  • z1=Ο‰=5βˆ’14+10+254iz_1 = \omega = \frac{\sqrt{5}-1}{4} + \frac{\sqrt{10+2\sqrt{5}}}{4}i
  • z2=Ο‰2=e4Ο€i/5=βˆ’5βˆ’14+10βˆ’254iz_2 = \omega^2 = e^{4\pi i/5} = \frac{-\sqrt{5}-1}{4} + \frac{\sqrt{10-2\sqrt{5}}}{4}i
  • z3=Ο‰3=e6Ο€i/5=βˆ’5βˆ’14βˆ’10βˆ’254iz_3 = \omega^3 = e^{6\pi i/5} = \frac{-\sqrt{5}-1}{4} - \frac{\sqrt{10-2\sqrt{5}}}{4}i
  • z4=Ο‰4=e8Ο€i/5=5βˆ’14βˆ’10+254iz_4 = \omega^4 = e^{8\pi i/5} = \frac{\sqrt{5}-1}{4} - \frac{\sqrt{10+2\sqrt{5}}}{4}i

Properties: These form a regular pentagon on the unit circle. They satisfy z5=1z^5 = 1 and 1+Ο‰+Ο‰2+Ο‰3+Ο‰4=01 + \omega + \omega^2 + \omega^3 + \omega^4 = 0.

Problem 4: Modulus Inequalities

Prove that ∣z1+z2βˆ£β‰€βˆ£z1∣+∣z2∣|z_1 + z_2| \leq |z_1| + |z_2| for all z1,z2∈Cz_1, z_2 \in \mathbb{C}.

Solution

Step 1: Compute ∣z1+z2∣2=(z1+z2)(z1+z2)β€Ύ=(z1+z2)(z1β€Ύ+z2β€Ύ)|z_1 + z_2|^2 = (z_1 + z_2)\overline{(z_1 + z_2)} = (z_1 + z_2)(\overline{z_1} + \overline{z_2})

=z1z1β€Ύ+z1z2β€Ύ+z2z1β€Ύ+z2z2β€Ύ=∣z1∣2+z1z2β€Ύ+z1z2β€Ύβ€Ύ+∣z2∣2= z_1\overline{z_1} + z_1\overline{z_2} + z_2\overline{z_1} + z_2\overline{z_2} = |z_1|^2 + z_1\overline{z_2} + \overline{z_1\overline{z_2}} + |z_2|^2

Step 2: Note z1z2β€Ύ+z1z2β€Ύβ€Ύ=2Re⁑(z1z2β€Ύ)z_1\overline{z_2} + \overline{z_1\overline{z_2}} = 2\operatorname{Re}(z_1\overline{z_2}).

So ∣z1+z2∣2=∣z1∣2+2Re⁑(z1z2β€Ύ)+∣z2∣2|z_1 + z_2|^2 = |z_1|^2 + 2\operatorname{Re}(z_1\overline{z_2}) + |z_2|^2.

Step 3: Since Re⁑(w)β‰€βˆ£w∣\operatorname{Re}(w) \leq |w| for any ww, we have 2Re⁑(z1z2β€Ύ)≀2∣z1∣∣z2β€Ύβˆ£=2∣z1∣∣z2∣2\operatorname{Re}(z_1\overline{z_2}) \leq 2|z_1||\overline{z_2}| = 2|z_1||z_2|.

Step 4: Therefore ∣z1+z2∣2β‰€βˆ£z1∣2+2∣z1∣∣z2∣+∣z2∣2=(∣z1∣+∣z2∣)2|z_1 + z_2|^2 \leq |z_1|^2 + 2|z_1||z_2| + |z_2|^2 = (|z_1| + |z_2|)^2.

Taking square roots: ∣z1+z2βˆ£β‰€βˆ£z1∣+∣z2∣|z_1 + z_2| \leq |z_1| + |z_2|. β–‘\square

Problem 5: Complex Equation Solving

Solve z2+2z+5=0z^2 + 2z + 5 = 0 for z∈Cz \in \mathbb{C}.

Solution

Step 1: Apply the quadratic formula: z=βˆ’2Β±4βˆ’202=βˆ’2Β±βˆ’162=βˆ’2Β±4i2z = \frac{-2 \pm \sqrt{4 - 20}}{2} = \frac{-2 \pm \sqrt{-16}}{2} = \frac{-2 \pm 4i}{2}

Step 2: z1=βˆ’1+2iz_1 = -1 + 2i and z2=βˆ’1βˆ’2iz_2 = -1 - 2i

Verification: (z+1)2=z2+2z+1(z+1)^2 = z^2 + 2z + 1, so z2+2z+5=(z+1)2+4=0z^2 + 2z + 5 = (z+1)^2 + 4 = 0 implies (z+1)2=βˆ’4(z+1)^2 = -4, so z+1=Β±2iz+1 = \pm 2i, giving z=βˆ’1Β±2iz = -1 \pm 2i. βœ“

Geometric interpretation: These roots are reflections of each other across the real axis (they are complex conjugates). They lie at distance 1+4=5\sqrt{1+4} = \sqrt{5} from the origin.


Common Mistakes

MistakeCorrectionExample
Forgetting that arg⁑(z)\arg(z) is multi-valuedThe principal argument Arg⁑(z)∈(βˆ’Ο€,Ο€]\operatorname{Arg}(z) \in (-\pi, \pi], but arg⁑(z)=Arg⁑(z)+2kΟ€\arg(z) = \operatorname{Arg}(z) + 2k\piarg⁑(βˆ’1)=Ο€+2kΟ€\arg(-1) = \pi + 2k\pi
Multiplying conjugates incorrectlyz1z2β€Ύ=z1β€Ύβ‹…z2β€Ύ\overline{z_1 z_2} = \overline{z_1} \cdot \overline{z_2}, NOT z1β€Ύ+z2β€Ύ\overline{z_1} + \overline{z_2}(2+3i)(1+i)β€Ύ=(2βˆ’3i)(1βˆ’i)\overline{(2+3i)(1+i)} = (2-3i)(1-i)
Confusing ∣z∣2|z|^2 with z2z^2∣z∣2=zzβ€Ύ=a2+b2|z|^2 = z\overline{z} = a^2+b^2 (real), while z2=a2βˆ’b2+2abiz^2 = a^2 - b^2 + 2abi (complex)∣3+4i∣2=25β‰ (3+4i)2=βˆ’7+24i|3+4i|^2 = 25 \neq (3+4i)^2 = -7+24i
Wrong quadrant for argumentUse atan2⁑(y,x)\operatorname{atan2}(y, x), not just arctan⁑(y/x)\arctan(y/x)arg⁑(βˆ’1βˆ’i)=βˆ’3Ο€/4\arg(-1-i) = -3\pi/4, not Ο€/4\pi/4
Assuming eiΞΈ=cos⁑θ+sin⁑θe^{i\theta} = \cos\theta + \sin\thetaIt's eiΞΈ=cos⁑θ+isin⁑θe^{i\theta} = \cos\theta + i\sin\theta β€” don't forget the iieiΟ€/2=ie^{i\pi/2} = i, not 11
Applying De Moivre to negative rrDe Moivre requires r>0r > 0; rewrite negative modulus first(βˆ’2eiΞΈ)n=2nein(ΞΈ+Ο€)(-2e^{i\theta})^n = 2^n e^{in(\theta + \pi)}
Forgetting that roots come in conjugate pairsIf z0z_0 is a root of a real polynomial, so is z0β€Ύ\overline{z_0}If 2+i2+i is a root, so is 2βˆ’i2-i

Interview / Exam Questions

Q1: What is Euler's formula, and why is it significant?

A1: Euler's formula states eiΞΈ=cos⁑θ+isin⁑θe^{i\theta} = \cos\theta + i\sin\theta. Its significance is threefold: (1) it unifies exponential and trigonometric functions, (2) it provides a compact representation of rotations in the complex plane (eiΞΈe^{i\theta} rotates a point by angle ΞΈ\theta), and (3) it yields Euler's identity eiΟ€+1=0e^{i\pi} + 1 = 0, connecting five fundamental constants. It is the foundation of Fourier analysis, phasor representation in electrical engineering, and much of complex analysis.


Q2: Why can't ∣z∣2|z|^2 equal z2z^2 for a nonzero complex number?

A2: ∣z∣2=a2+b2|z|^2 = a^2 + b^2 is always a non-negative real number, while z2=a2βˆ’b2+2abiz^2 = a^2 - b^2 + 2abi is real only if ab=0ab = 0. For z=1+iz = 1+i: ∣z∣2=2|z|^2 = 2 but z2=2iz^2 = 2i. The modulus squared is a geometric quantity (distance squared from origin), while z2z^2 is an algebraic operation (squaring the complex number).


Q3: What are the nn-th roots of unity, and what geometric figure do they form?

A3: The nn-th roots of unity are the solutions zn=1z^n = 1, given by zk=e2Ο€ik/nz_k = e^{2\pi ik/n} for k=0,1,…,nβˆ’1k = 0, 1, \ldots, n-1. They are equally spaced on the unit circle at angles 0,2Ο€/n,4Ο€/n,…,2Ο€(nβˆ’1)/n0, 2\pi/n, 4\pi/n, \ldots, 2\pi(n-1)/n. Geometrically, they are the vertices of a regular nn-gon inscribed in the unit circle. For n=3n = 3 they form an equilateral triangle; for n=4n = 4, a square; for n=6n = 6, a regular hexagon.


Q4: If z1z2=0z_1 z_2 = 0, must z1=0z_1 = 0 or z2=0z_2 = 0? Prove or disprove.

A4: Yes. If z1z2=0z_1 z_2 = 0, then ∣z1z2∣=∣z1βˆ£β‹…βˆ£z2∣=0|z_1 z_2| = |z_1| \cdot |z_2| = 0. Since ∣z1∣|z_1| and ∣z2∣|z_2| are non-negative reals, one of them must be zero. If ∣z1∣=0|z_1| = 0 then z1=0z_1 = 0 (since ∣z∣=0|z| = 0 iff z=0z = 0). Similarly for z2z_2. This is the zero-product property, which holds in C\mathbb{C} just as in R\mathbb{R}.


Q5: How do you compute arg⁑(z1/z2)\arg(z_1 / z_2) and what subtleties arise?

A5: arg⁑(z1/z2)=arg⁑(z1)βˆ’arg⁑(z2)(mod2Ο€)\arg(z_1/z_2) = \arg(z_1) - \arg(z_2) \pmod{2\pi}. The subtlety is that if you use principal arguments, the result may not be a principal argument. For example, Arg⁑(βˆ’1)=Ο€\operatorname{Arg}(-1) = \pi and Arg⁑(i)=Ο€/2\operatorname{Arg}(i) = \pi/2, but Arg⁑(βˆ’1/i)=Arg⁑(i)=Ο€/2β‰ Ο€βˆ’Ο€/2=Ο€/2\operatorname{Arg}(-1/i) = \operatorname{Arg}(i) = \pi/2 \neq \pi - \pi/2 = \pi/2 (happens to work here). But Arg⁑((βˆ’1)/(1+i))\operatorname{Arg}((-1)/(1+i)) needs adjustment. Always reduce modulo 2Ο€2\pi to the interval (βˆ’Ο€,Ο€](-\pi, \pi].


Q6: Prove that if ∣z∣=1|z| = 1, then zβ€Ύ=1/z\overline{z} = 1/z.

A6: If ∣z∣=1|z| = 1, then zzβ€Ύ=∣z∣2=1z\overline{z} = |z|^2 = 1. Dividing both sides by zz (which is nonzero since ∣z∣=1|z|=1): zβ€Ύ=1/z\overline{z} = 1/z. Geometrically, this means the conjugate of a point on the unit circle is its reciprocal, which is also on the unit circle.


Quick Reference

Formula Summary

FormulaExpressionNotes
Modulusβˆ₯zβˆ₯=a2+b2\|z\| = \sqrt{a^2 + b^2}Distance from origin
Conjugatea+biβ€Ύ=aβˆ’bi\overline{a+bi} = a - biReflection across real axis
Polar Formz=reiΞΈz = re^{i\theta}r=βˆ₯zβˆ₯r = \|z\|, ΞΈ=arg⁑(z)\theta = \arg(z)
Euler's Formulaeiθ=cos⁑θ+isin⁑θe^{i\theta} = \cos\theta + i\sin\thetaFundamental identity
De Moivre(reiΞΈ)n=rneinΞΈ(re^{i\theta})^n = r^n e^{in\theta}Powers in polar form
N-th Rootz1/n=r1/nei(ΞΈ+2kΟ€)/nz^{1/n} = r^{1/n}e^{i(\theta+2k\pi)/n}nn distinct roots
Productz1z2=r1r2ei(ΞΈ1+ΞΈ2)z_1z_2 = r_1r_2 e^{i(\theta_1+\theta_2)}Multiply moduli, add arguments
Quotientz1/z2=(r1/r2)ei(ΞΈ1βˆ’ΞΈ2)z_1/z_2 = (r_1/r_2)e^{i(\theta_1-\theta_2)}Divide moduli, subtract arguments
Triangle Inequalityβˆ₯z1+z2βˆ₯≀βˆ₯z1βˆ₯+βˆ₯z2βˆ₯\|z_1+z_2\| \leq \|z_1\| + \|z_2\|Equality iff z1,z2z_1, z_2 are collinear
Modulus Squaredzzβ€Ύ=βˆ₯zβˆ₯2z\overline{z} = \|z\|^2Always real, non-negative
Cosine Formulacos⁑θ=(eiΞΈ+eβˆ’iΞΈ)/2\cos\theta = (e^{i\theta} + e^{-i\theta})/2From Euler's formula
Sine Formulasin⁑θ=(eiΞΈβˆ’eβˆ’iΞΈ)/(2i)\sin\theta = (e^{i\theta} - e^{-i\theta})/(2i)From Euler's formula

Cross-References

  • 092 - Complex Functions β€” Analyticity, Cauchy-Riemann equations, and conformal mappings build on the algebraic foundations of complex numbers.
  • 093 - Contour Integration β€” Contour integrals use polar form and De Moivre's theorem to parameterize paths in the complex plane.
  • 094 - Residue Theory β€” Finding poles and computing residues requires fluency with complex arithmetic and roots of unity.
  • 095 - Applications β€” Signal processing (Fourier transforms) and control theory (Z-transforms) use complex numbers as their fundamental language.
  • Linear Algebra (Topic 14) β€” Eigenvalues of real matrices may be complex; the characteristic polynomial roots live in C\mathbb{C}.
  • Differential Equations β€” Complex exponentials e(a+bi)te^{(a+bi)t} arise in solutions to linear ODEs with complex characteristic roots.
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