2.4 Complex numbers
No real number squares to give a negative, so \(x^2=-1\) has no real solution. Rather than stop there, we give that missing number a name.
Definition 2.12. The imaginary unit \(i\) is defined by \[i^2=-1,\qquad \text {that is}\qquad i=\sqrt {-1}.\] A complex number has the form \(z=a+bi\) with \(a,b\in \mathbb {R}\). Here \(a\) is the real part and \(b\) the imaginary part, written \(\operatorname {Re}(z)=a\) and \(\operatorname {Im}(z)=b\).
Note 2.13. The imaginary part is the real number \(b\), not \(bi\). For \(z=3+4i\), \(\operatorname {Im}(z)=4\).
Two complex numbers are equal exactly when their real parts match and their imaginary parts match. That single fact turns one complex equation into two real ones, which is how such equations are usually solved.
2.4.1 Arithmetic with complex numbers
Addition and subtraction are done part by part. Multiplication is ordinary expansion, followed by replacing \(i^2\) with \(-1\).
Example 2.14. Let \(z_1=3+4i\) and \(z_2=1-2i\). Find (a) \(z_1+z_2\), (b) \(z_1-z_2\), (c) \(z_1z_2\).
Solution. (a) Add real to real, imaginary to imaginary: \[z_1+z_2=(3+1)+(4-2)i=4+2i.\]
(b) \[z_1-z_2=(3-1)+(4-(-2))i=2+6i.\]
(c) Expand as usual: \begin {align*} z_1z_2 &=(3+4i)(1-2i)\\ &=3-6i+4i-8i^2\\ &=3-2i-8i^2. \end {align*}
Now use \(i^2=-1\), so \(-8i^2=-8(-1)=+8\): \[z_1z_2=3+8-2i=11-2i.\]
Definition 2.15. The conjugate of \(z=a+bi\) is \(\bar {z}=a-bi\): the same number with the sign of the imaginary part reversed.
The conjugate matters because a number times its conjugate is always real: \[(a+bi)(a-bi)=a^2-(bi)^2=a^2-b^2i^2=a^2+b^2.\] That is exactly the tool needed for division — the same idea as rationalising a denominator.
Solution. Multiply top and bottom by the conjugate of the denominator, \(1+2i\): \[\frac {3+4i}{1-2i}\times \frac {1+2i}{1+2i}=\frac {(3+4i)(1+2i)}{1^2+2^2}.\]
The denominator is now real: \(1^2+2^2=5\). Expanding the numerator, \[(3+4i)(1+2i)=3+6i+4i+8i^2=3+10i-8=-5+10i.\] \[\therefore \quad \frac {3+4i}{1-2i}=\frac {-5+10i}{5}=-1+2i.\]
Check. Multiply back: \((-1+2i)(1-2i)=-1+2i+2i-4i^2=-1+4i+4=3+4i\), which is the original numerator. Any division can be checked this way.
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