Solved Exercises on Complex Numbers

Exercise editors and source: Prof. Antonino De Martino and Dr. Luana Manfredini, Eserciziario 2.1. The original exercises and source theory are presented here in English for Logic & Motion.

This collection develops the main techniques used to work with complex numbers, from algebraic and exponential representations to modulus, conjugation, powers, roots, and equations in the complex plane.

The theoretical recalls below provide the formulas and concepts required for the solved exercises.

Theoretical Recalls

  • Complex numbers are defined as ordered pairs of real numbers, with algebraic form
\[z = a + ib, \quad i^2 = -1 .\]
  • Conjugate and modulus:
\[\overline{z} = a - ib, \quad |z| = \sqrt{a^2+b^2} .\]
  • Polar and exponential form: for z ≠ 0, the argument θ is defined modulo 2π. Zero has modulus zero but no defined argument.
\[z = r(\cos \theta + i\sin \theta) = re^{i\theta}, \quad r = |z|, \ \theta = \arg(z) .\]
  • De Moivre’s formula: n is an integer; for negative n, z must be nonzero.
\[(re^{i\theta})^n = r^n e^{in\theta} .\]
  • n-th roots of unity:
\[z_k = e^{\tfrac{2k\pi i}{n}}, \quad k=0,1,\dots,n-1 .\]
  • Useful relations:
\[|zw| = |z|\,|w|, \quad \overline{zw} = \overline{z}\,\overline{w} .\]

Theoretical Recalls

The idea of complex numbers began to emerge around the 16th century, when Italian algebraists attempted to solve algebraic equations.
Rafael Bombelli’s L’Algebra (1572) gave systematic rules for calculating with square roots of negative numbers. The designation imaginary appears later in Descartes’ La Géométrie (1637). Their geometric interpretation was developed further in the late eighteenth and early nineteenth centuries.
One of the most important theorems based on complex numbers is the Fundamental Theorem of Algebra.

Given z = x + iy ∈ ℂ, the most important formulas are:

  • Modulus
\[|z| = \sqrt{x^{2}+y^{2}} .\]
  • Trigonometric form
\[z = |z|(\cos \theta + i\sin \theta) .\]
  • Euler’s formula
\[e^{i\theta} = \cos\theta + i\sin\theta .\]
  • Exponential form
\[z = |z|e^{i\theta} .\]
  • De Moivre’s formula
\[z^{n} = \rho^{n}(\cos(n\theta)+i\sin(n\theta)) .\]
  • n-th roots of a complex number: let n be a positive integer, ρ = z > 0, and θ an argument of z. The n roots are distinct. If z = 0, the only root is 0.
\[w_{k} = \sqrt[n]{\rho}\Bigl[\cos\!\Bigl(\tfrac{\theta+2k\pi}{n}\Bigr) + i\sin\!\Bigl(\tfrac{\theta+2k\pi}{n}\Bigr)\Bigr], \quad k=0,1,\dots,n-1 .\]

Exercises

Exercise 1 — Sixth Roots of −1

Solution. Since −1 = 1 and arg(−1) = π, by the n-th roots formula:
\[w_k=\sqrt[6]{1}\Bigl[\cos\!\Bigl(\tfrac{\pi+2k\pi}{6}\Bigr)+i\sin\!\Bigl(\tfrac{\pi+2k\pi}{6}\Bigr)\Bigr],\quad k=0,\dots,5 .\]

Thus

\[\frac{\sqrt{3}}{2}+\frac{i}{2},\quad i,\quad -\frac{\sqrt{3}}{2}+\frac{i}{2},\quad -\frac{\sqrt{3}}{2}-\frac{i}{2},\quad -i,\quad \frac{\sqrt{3}}{2}-\frac{i}{2}.\]

Final result

\[w=\Bigl\{\tfrac{\sqrt{3}}{2}\pm\tfrac{i}{2},\ \pm i,\ -\tfrac{\sqrt{3}}{2}\pm\tfrac{i}{2}\Bigr\}.\]

Exercise 2 — Solve |z|z̄ = 2i

Solution. Taking moduli:

\[||z|\,\overline{z}|=|2i| \ \Rightarrow\ |z|^2=2 \ \Rightarrow\ |z|=\sqrt{2} .\]

Back to the equation:

\[\sqrt{2}\,\overline{z}=2i \ \Rightarrow\ \overline{z}=2i/\sqrt{2} \ \Rightarrow\ z=-\sqrt{2}\,i .\]

Final result

\[z=-\sqrt{2}\,i .\]

Exercise 3 — Solve the System

\[\begin{cases} z^{2}+w^{4}=0 \\[6pt] z^{3}\,\overline{w}^{\,5}=1 \end{cases}\]

Solution. Squaring the second equation gives a necessary condition. The resulting system may have extra solutions, so candidates must be checked in the original equations:

\[\begin{cases} z^{2}=-w^{4} \\[6pt] z^{6}\,\overline{w}^{\,10}=1 \end{cases}\]

From the first: z⁶ = −w¹². Substitute:

\[w^{12}\,\overline{w}^{\,10}=-1 \ \Rightarrow\ w^{2}|w|^{20}=-1 .\]

Taking moduli: |w|²² = 1 ⇒ |w| = 1. Then w² = −1 ⇒ w = ±i. From z² = −1 ⇒ z = ±i.
Valid solutions: (i, −i) and (−i, i).

Final result

\[(z,w)\in\{(i,-i),\ (-i,i)\} .\]

Exercise 4 — Solve z² + (i − 1)z − i = 0

Solution. Discriminant:

\[\Delta=(i-1)^{2}+4i=1+2i+i^{2}=(1+i)^{2} .\]

Thus z₁ = 1, z₂ = −i.
Factorization check: (z − 1)(z + i) = 0.

Final result

\[z\in\{1,\ -i\}.\]

Exercise 5 — Solve z⁴ = z̄³

Solution. Multiply by z³:

\[z^{7}=|z|^{6} .\]
Taking moduli: z ⁴ = z ³ ⇒ z = 0 or z = 1.
If z = 1, then z⁷ = 1 ⇒ z are 7th roots of unity.        

Final result

\[z=0 \quad \text{or} \quad z=e^{2\pi i k/7},\ k=0,\dots,6 .\]

Exercise 6 — Solve z² + zz̄ = 1 + i

Solution. Let z = x + iy:

\[(x+iy)^{2}+(x+iy)(x-iy)=1+i\] \[2x^{2}+2ixy=1+i .\]

So x² = ½, xy = ½ ⇒ x = ±√2/2, y = ±√2/2 with same sign.

Final result

\[z\in\Bigl\{\tfrac{1+i}{\sqrt{2}},\ -\tfrac{1+i}{\sqrt{2}}\Bigr\}.\]

Exercise 7 — Solve Re(z²) = z + i

Solution. Re(z²) = x² − y². Equation:

\[x^{2}-y^{2}=x+i(y+1) .\]

So y = −1 and x² − (−1)² = x ⇒ x² − 1 = x ⇒ x = (1 ± √5)/2.

Final result

\[z=\tfrac{1\pm\sqrt{5}}{2}-i .\]

Exercise 8 — Solve i(z + 1)³ = i

Solution. Divide by i: (z + 1)³ = 1. Let ω = z + 1 ⇒ ω³ = 1 ⇒ ω ∈ {1, −½ ± i√3/2}.
So z ∈ {0, −3/2 ± i√3/2}.

Final result

\[z\in\Bigl\{0,\ -\tfrac{3}{2}\pm i\tfrac{\sqrt{3}}{2}\Bigr\}.\]

Exercise 9 — Solve (z⁵z̄² − 1)(z² + z + 2) = 0

Solution. Case 1: z⁵z̄² = 1. Taking moduli gives |z|⁷ = 1, hence |z| = 1. Thus z̄ = 1/z and z⁵z̄² = z³, so z³ = 1 and z ∈ {1, −½ ± i√3/2}.
Case 2: z² + z + 2 = 0 ⇒ z = (−1 ± i√7)/2.

Final result

\[z\in\Bigl\{1,\ -\tfrac{1}{2}\pm i\tfrac{\sqrt{3}}{2},\ \tfrac{-1\pm i\sqrt{7}}{2}\Bigr\}.\]

Exercise 10 — Solve (2z − 1)²(2z̄ + 1) = 4z(2z − 1)

Solution. Either 2z − 1 = 0 ⇒ z = ½, or:

\[(2z-1)(2\overline{z}+1)=4z .\]

Let z = x + iy:

\[x^{2}+y^{2}-x=\tfrac{1}{4} .\]

This is a circle centered at (½, 0), with radius √2/2. The solution z = ½ is an additional isolated point: it is the center of the circle and does not lie on its circumference.

Final result

\[z=\tfrac{1}{2}\ \ \text{or}\ \ x^{2}+y^{2}-x=\tfrac{1}{4}.\]

Complex numbers connect algebraic equations with geometry, trigonometry, and exponential representations.

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