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.