Background
Imaginary numbers are of the form $ a+bi $, where $ a $ is the real part, $ b $ is the imaginary part, and $ i $ is a number with the property that $ i^2=-1 $. As Electrical Engineers, we substitute $ j $ for $ i $ because $ i $ is usually reserved for current; this ideally makes things less confusing.
Complex numbers are used to simplify analysis of circuits that would otherwise require us to write high-order differential equations.
Operations with Complex Numbers
Raising $ j $ to Powers
Since we know that $ j=\sqrt{-1} $, we can deduce the following cyclic pattern:
$ j^{4n}=j $
$ j^{4n+1}=-1 $
$ j^{4n+2}=-j $
$ j^{4n+3}=1 $
From this we see that the pattern $ \{j, -1, -j, 1\} $ repeats as $ j $ is raised to higher and higher powers.
Addition and Subtraction
To add or subtract complex numbers, simply add them component-wise as you would a vector.
- $ (1+2j)+(3-8j) = (1+3)+(2-8)j=4-6j $
- $ (1-2j)-(3-8j)=(1-3)+(-2-(-8))j=-2+6j $
Conjugation
The complex conjugate of a complex number, denoted by placing a bar over the imaginary number, is obtained by simply changing the sign on the imaginary part.
- $ \overline{4j}=-4j $
- $ \overline{1+2j}=1-2j $
- $ \overline{3-8j}=3+8j $
Multiplication
To multiply complex numbers, use the FOIL method of binomial multiplication. This has the effect of multiplying each element of the first term by each element of the second term, then adding all the intermediate results.
- $ (1+2j)(3-8j)=1*3+1*(-8j)+2j*3+2j*(-8j)=3-8j+6j-16j^2=19-2j $
- $ (2-j)(1+j)=2*1+2j-j-j^2=3+j $
Multiplying a complex number by its conjugate is equivalent to adding the squares of the coefficients of its real and imaginary parts.
Division
Division of two complex numbers uses the conjugate to give a real-valued divisor. Consider the example $ \frac{1+2j}{3+8j} $.
First multiply both numerator and denominator by the complex conjugate of the denominator. This will give a real-valued denominator. Performing this operating yields $ \frac{(1+2j)(3-8j)}{(3+8j)(3-8j)} $ which gives us $ \frac{19-2j}{64} $, because multiplying a complex number by its conjugate gives the square of the norm of that complex number.
The resulting number can then be simplified. In this case, we have $ \frac{19}{64}-\frac{j}{32} $.
