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Complex Numbers

Introduction

Just as we can perform operations with real numbers, we can also add, subtract, multiply, and divide complex numbers. Recall the order of operations for real numbers here: Operations with Numbers

When performing operations with complex numbers, you may encounter three types of complex numbers:

  • Real number as complex number - imaginary part is equal to 0

Examples: \(2 ,  -6,  7.462437,  \frac{3}{5}\)

Tip: Since all real numbers are also complex numbers, you can add \(0i\) to these numbers and they'll still be considered complex numbers!

  • Imaginary number as complex number - real part is equal to 0

Examples: \(i,  3i,  \sqrt{2}i,  \frac{1}{2}i\)

  • Complex number composed of real and imaginary part

Examples: \(3+4i,  \frac{5}{2}-0.1i,  -7.25-11i\)

Addition/Subtraction

To add or subtract complex numbers, just combine like terms as if the number \(i\) is a variable, then evaluate.

  • For any two complex numbers, say \(x=a+bi\) and \(y=c+di\), we can add them as follows:

\(x+y=(a+bi)+(c+di)=(a+c)+(b+d)i\)

  • For any two complex numbers, say \(x=a+bi\) and \(y=c+di\), we can subtract them as follows:

\(x-y=(a+bi)-(c+di)=(a-c)+(b-d)i\)

Example: Evaluate \((-4+5i)-(6-6i)+(2i+7)\).

Solution:

1. Combine like terms:

\((-4+5i)-(6-6i)+(2i+7)\)

\(=(-4-6+7)+(5i-(-6i)+2i)\)

2. Simplify and evaluate:

\(=-3+(5-(-6)+2)i\)

\(=-3+13i\)

Multiplication

To multiply complex numbers, you can first use one the FOIL process to expand the two binomials. The FOIL process is a quick method of apply the distributive law to expand two binomials:

\((a+b)(c+d)=ac+ad+bc+bd\)

where FOIL is an acronym for:

  • First
  • Outer
  • Inner
  • Last

since we multiply the "firsts" in the two binomials - \(a\) and \(c\) in the above case, then the "outer" terms - \(a\) and \(d\), then "inner" terms - \(b\) and \(c\), then "last" terms - \(b\) and \(d\), and add all these products up.

Once you've finished expanding, the final step is to combine like terms. To summarize, you can multiply any two complex numbers, say, \((a+bi)\) and \((c+di)\), as follows:

\((a+bi)(c+di)=(a)(c)+(a)(di)+(bi)(c)+(bi)(di)=ac+adi+bci+bd(-1)=(ac-bd)+(ad+bc)i\)

*Note: in the second last step, we were able to replace \(i\times i=i^2\) with -1 since, by definition, \(i^2=(\sqrt{-1})^2=-1\)

Example: Evaluate \((3+7i)(2-i)\).

Solution:

1. First, expand using the FOIL method:

\((3+7i)(2-i)\)

\(=(3)(2)+(3)(-i)+(7i)(2)+(7i)(-i)\)

2. Next, multiply out all the terms:

\(=6-3i+14i-7i^2\)

\(=6-3i+14i-7(-1)\)

\(=6-3i+14i+7\)

3. Evaluate by collecting like terms and simplifying:

\(=13+11i\)

Division

Complex division is often represented using fractions, in the general form of \(\frac{a+bi}{c+di}\) for any two complex numbers. Notice that the denominator is the sum of two terms, so we cannot find the quotient directly. 

Recall that the conjugate of a complex number \(z=a+bi\) is: \(\bar{z}=a-bi\). This will be helpful to us when dividing complex numbers since multiplying a complex number by its conjugate gives us a real number product.

For any two complex numbers, say \(x=a+bi\) and \(y=c+di\), we can divide \(x\) by \(y\)  (i.e. evaluate \(\frac{a+bi}{c+di}\)) by following these steps:

1. Determine the conjugate of the denominator (which is \(c-di\) here). Then multiply the numerator and denominator by this conjugate:

\(\frac{a+bi}{c+di} \cdot \frac{c-di}{c-di}\)

*Note: this step does not change the original expression since we're actually just multiplying it by \(\frac{c-di}{c-di}=1\)! 

2. Expand the expression by multiplying:

\(=\frac{(a+bi)(c-di)}{(c+di)(c-di)}\)

\(=\frac{ac-adi+bci+bd}{c^2-d^2i^2}\)

3. Simplify by collecting like terms and reducing fractions (if applicable):

\(=\frac{(ac-bd)+(bc-ad)i}{c^2+d^2}\)

 

Example: Evaluate \(\frac{3+5i}{2+4i}\).

See the below video for the solution:

 

Practice

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