Why?

Complex Numbers make it possible to solve square roots of negative numbers and for engineers, they make it easier to solve complex differential equations. Usually complex numbers are of the form $ a+bi $,where a is the real part of the number and b is the imaginary part with $ i=\sqrt{-1} $, but using i is arbitrary. it can be anything. i is convenient because imaginary begins with i, but we will mostly use j to denote $ \sqrt{-1} $

Properties

$ j^{4n}=j $

$ j^{4n+1}=-1 $

$ j^{4n+2}=-j $

$ j^{4n+3}=1 $

where n is a whole number.

Ex:

$ j^9=j^{4*2+1}=j^8*j^1=1*j=j+ $

Addition

$ (a+bj)+(c+dj)=(a+c)+(b+d)j $

Ex:

$ (7+3j)+(4+5j)=(7+4)+(3+5)j=11+8j $ $ (7+3j)-(4+5j)=(7-4)+(3-5)j=3-2j $

Multiplication

$ (a+bj)*(c+dj)=a*c+a*dj+bj*c+bj*dj=(a*c-b*d)+(a*d+b*c)j $

Ex:

$ (2+3j)*(4+5j)=2*4+2*5j+3j*4+3j*5j=(2*4-3*5)+(2*5+3*4)j=-7+22j $

Conjugate

Suppose $ n=a+bj $, then the conjugate of n, m, is defined as $ m=a-bj $. Note: When multiplying conjugate pairs, the result is no longer an imaginary number. $ (a+bj)*(a-bj)=a*a-a*bj+bj*a+b^2=a^2+b^2 $

Euler's Formula

This formula is very useful when working with signals.

$ e^{wjt}=cos{wt}+jsin{wt} $

Alumni Liaison

Recent Math PhD now doing a post-doctorate at UC Riverside.

Kuei-Nuan Lin