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:
\[z = r(\cos \theta + i\sin \theta) = re^{i\theta}, \quad r = |z|, \ \theta = \arg(z) .\]
  • De Moivre’s formula:
\[(re^{i\theta})^n = r^n e^{in\theta} .\]
  • 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.
The term imaginary number was introduced by Bombelli (1572), while a more rigorous foundation was given by Gauss in the early 19th century.
One of the most important theorems based on complex numbers is the Fundamental Theorem of Algebra.

Given z = x + i y ∈ 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
\[w_{k} = \sqrt[n]{\rho}\Bigl[\cos\!\Bigl(\tfrac{\theta+2k\pi}{n}\Bigr) + i\sin\!\Bigl(\tfrac{\theta+2k\pi}{n}\Bigr)\Bigr], \quad k=0,1,\dots,n-1 .\]

Exercises

Ex 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

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

Solution. Taking moduli:

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

Back to the equation:

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

Final result

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

Ex 3. Solve the system

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

Solution. Equivalent system:

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

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

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

Ex 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

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

Ex 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

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

Ex 6. Solve z²+z z̄ = 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

\[z\in\Bigl\{\tfrac{\sqrt{2}}{2}\pm i\tfrac{\sqrt{2}}{2}\Bigr\}.\]

Ex 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

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

Ex 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

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

Ex 9. Solve (z⁵ z̄² − 1)(z²+z+2)=0.

Solution. Case 1: |z|=1 ⇒ z³=1 ⇒ z∈{1, −½±i√3/2}.
Case 2: z²+z+2=0 ⇒ z=(−1±i√7)/2.

Final result

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

Ex 10. Solve (2z−1)²(2z̄+1)=4z(2z−1).

Solution. Either 2z−1=0 ⇒ z=½, or:

\[(2z-1)(2z̄+1)=4z .\]

Let z=x+iy:

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

That is circle centered at (½,0), radius √2/2, including z=½.

Final result

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

Explore more topics in Calculus 1