Complex numbers extend the real number system so equations like \(x^2+1=0\) can be solved. In high school, you usually write a complex number in Cartesian form: \[ z = a + bi, \] where \(a,b\in\mathbb{R}\) and \(i^2=-1\).
Definition
\[ i^2 = -1 \]
Useful cycle: \[ i^1=i,\quad i^2=-1,\quad i^3=-i,\quad i^4=1, \] and then the pattern repeats every 4 powers.
For \(z=a+bi\):
Example. If \(z=3-4i\), then \(\Re(z)=3\), \(\Im(z)=-4\).
Add/subtract by combining like terms:
\[ (a+bi)\pm(c+di)=(a\pm c)+(b\pm d)i. \]
Multiply using expansion and \(i^2=-1\):
\[ (a+bi)(c+di)=ac+adi+bci+bd\,i^2=(ac-bd)+(ad+bc)i. \]
Example. \((2+3i)(1-4i)=2-8i+3i-12i^2=14-5i.\)
The conjugate of \(z=a+bi\) is \[ \overline{z}=a-bi. \] Multiplying by conjugates removes imaginary parts: \[ z\overline{z}=(a+bi)(a-bi)=a^2+b^2. \]
The modulus (distance from origin in Argand plane) is: \[ |z|=\sqrt{a^2+b^2}. \]
Example. For \(z=3-4i\), \(\overline{z}=3+4i\), \(|z|=5\).
To divide by \(c+di\), multiply top and bottom by \(c-di\):
\[ \frac{a+bi}{c+di} =\frac{(a+bi)(c-di)}{(c+di)(c-di)} =\frac{(a+bi)(c-di)}{c^2+d^2}. \]
Example. \(\dfrac{1+2i}{3-i}\):
\[ \frac{1+2i}{3-i}\cdot\frac{3+i}{3+i} =\frac{(1+2i)(3+i)}{10} =\frac{1+7i}{10} =\frac{1}{10}+\frac{7}{10}i. \]
Plot \(z=a+bi\) as the point \((a,b)\): horizontal axis is real part, vertical axis is imaginary part. This lets you interpret \(|z|\) as geometric distance.
If \(x^2+4=0\), then \(x^2=-4\), so \[ x=\pm 2i. \] Complex numbers ensure every quadratic has two roots (counting multiplicity), and later this extends to all polynomial equations.
Complex numbers first appeared when mathematicians tried to solve cubic equations in the 16th century. Algebraic formulas sometimes produced square roots of negative numbers, even when final answers were real.
Mathematicians like Cardano, Bombelli, and later Euler developed rules for these numbers. What began as a formal trick became a major mathematical system used in engineering, physics, and signal processing.
For \(c+di\): \[ (c+di)(c-di)=c^2-d^2i^2=c^2+d^2, \] which is real and positive (unless \(c=d=0\)). This is why multiplying by conjugates converts complex division into standard real-number division.
For \(z=a+bi\): \[ z\overline{z}=(a+bi)(a-bi)=a^2+b^2=|z|^2. \] So modulus is directly linked to conjugation and the Pythagorean distance in the Argand plane.
Pause and reason before continuing. Discuss with a classmate or write your reasoning in a notebook.
Checkpoint
Simplify \(i^{23}\) using the 4-step power cycle.
Checkpoint
Compute \((4-3i)+(2+5i)\) and \((4-3i)-(2+5i)\).
Checkpoint
Expand and simplify \((1+2i)(3-i)\) into \(a+bi\) form.
Checkpoint
Find \(\overline{z}\) and \(|z|\) when \(z=-5+12i\).
Checkpoint
Write \(\dfrac{2-i}{1+3i}\) in \(a+bi\) form.
Checkpoint
Solve \(x^2-6x+13=0\). Why are the roots complex?
Electrical engineers model alternating-current circuits with complex numbers because amplitude and phase are naturally represented together as one complex quantity.
Sound, radio, and image-processing systems use complex numbers to describe oscillations and frequency content efficiently (e.g., Fourier methods).
Multiplying by certain complex numbers corresponds to rotating and scaling points in the plane, making complex numbers useful in geometry and computer graphics.