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Plotting a complex number on the complex plane

We cannot plot complex numbers on a number line as we might real numbers. However, we can still represent them graphically. To represent a complex number we need to address the two components of the number. We use the complex plane    , which is a coordinate system in which the horizontal axis represents the real component and the vertical axis represents the imaginary component. Complex numbers are the points on the plane, expressed as ordered pairs ( a , b ) , where a represents the coordinate for the horizontal axis and b represents the coordinate for the vertical axis.

Let’s consider the number −2 + 3 i . The real part of the complex number is −2 and the imaginary part is 3 i . We plot the ordered pair ( −2 , 3 ) to represent the complex number −2 + 3 i as shown in [link] .

Plot of a complex number, -2 + 3i. Note that the real part (-2) is plotted on the x-axis and the imaginary part (3i) is plotted on the y-axis.

Complex plane

In the complex plane , the horizontal axis is the real axis, and the vertical axis is the imaginary axis as shown in [link] .

The complex plane showing that the horizontal axis (in the real plane, the x-axis) is known as the real axis and the vertical axis (in the real plane, the y-axis) is known as the imaginary axis.

Given a complex number, represent its components on the complex plane.

  1. Determine the real part and the imaginary part of the complex number.
  2. Move along the horizontal axis to show the real part of the number.
  3. Move parallel to the vertical axis to show the imaginary part of the number.
  4. Plot the point.

Plotting a complex number on the complex plane

Plot the complex number 3 4 i on the complex plane.

The real part of the complex number is 3 , and the imaginary part is −4 i . We plot the ordered pair ( 3 , −4 ) as shown in [link] .

Plot of a complex number, 3 - 4i. Note that the real part (3) is plotted on the x-axis and the imaginary part (-4i) is plotted on the y-axis.
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Plot the complex number −4 i on the complex plane.

Graph of the plotted point, -4-i.
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Adding and subtracting complex numbers

Just as with real numbers, we can perform arithmetic operations on complex numbers. To add or subtract complex numbers, we combine the real parts and combine the imaginary parts.

Complex numbers: addition and subtraction

Adding complex numbers:

( a + b i ) + ( c + d i ) = ( a + c ) + ( b + d ) i

Subtracting complex numbers:

( a + b i ) ( c + d i ) = ( a c ) + ( b d ) i

Given two complex numbers, find the sum or difference.

  1. Identify the real and imaginary parts of each number.
  2. Add or subtract the real parts.
  3. Add or subtract the imaginary parts.

Adding complex numbers

Add 3 4 i and 2 + 5 i .

We add the real parts and add the imaginary parts.

( a + b i ) + ( c + d i ) = ( a + c ) + ( b + d ) i ( 3 4 i ) + ( 2 + 5 i ) = ( 3 + 2 ) + ( 4 + 5 ) i                               = 5 + i
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Subtract 2 + 5 i from 3 4 i .

( 3 4 i ) ( 2 + 5 i ) = 1 9 i

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Multiplying complex numbers

Multiplying complex numbers is much like multiplying binomials. The major difference is that we work with the real and imaginary parts separately.

Multiplying a complex numbers by a real number

Let’s begin by multiplying a complex number by a real number. We distribute the real number just as we would with a binomial. So, for example,

Showing how distribution works for complex numbers. For 3(6+2i), 3 is multiplied to both the real and imaginary parts. So we have (3)(6)+(3)(2i) = 18 + 6i.

Given a complex number and a real number, multiply to find the product.

  1. Use the distributive property.
  2. Simplify.

Multiplying a complex number by a real number

Find the product 4 ( 2 + 5 i ) .

Distribute the 4.

4 ( 2 + 5 i ) = ( 4 2 ) + ( 4 5 i ) = 8 + 20 i
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Find the product 4 ( 2 + 6 i ) .

8 24 i

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Questions & Answers

So a horizontal compression by factor of 1/2 is the same as a horizontal stretch by a factor of 2, right?
KARMEL Reply
The center is at (3,4) a focus is at (3,-1), and the lenght of the major axis is 26
Rima Reply
The center is at (3,4) a focus is at (3,-1) and the lenght of the major axis is 26 what will be the answer?
Rima
I done know
Joe
What kind of answer is that😑?
Rima
I had just woken up when i got this message
Joe
Can you please help me. Tomorrow is the deadline of my assignment then I don't know how to solve that
Rima
i have a question.
Abdul
how do you find the real and complex roots of a polynomial?
Abdul
@abdul with delta maybe which is b(square)-4ac=result then the 1st root -b-radical delta over 2a and the 2nd root -b+radical delta over 2a. I am not sure if this was your question but check it up
Nare
This is the actual question: Find all roots(real and complex) of the polynomial f(x)=6x^3 + x^2 - 4x + 1
Abdul
@Nare please let me know if you can solve it.
Abdul
I have a question
juweeriya
hello guys I'm new here? will you happy with me
mustapha
The average annual population increase of a pack of wolves is 25.
Brittany Reply
how do you find the period of a sine graph
Imani Reply
Period =2π if there is a coefficient (b), just divide the coefficient by 2π to get the new period
Am
if not then how would I find it from a graph
Imani
by looking at the graph, find the distance between two consecutive maximum points (the highest points of the wave). so if the top of one wave is at point A (1,2) and the next top of the wave is at point B (6,2), then the period is 5, the difference of the x-coordinates.
Am
you could also do it with two consecutive minimum points or x-intercepts
Am
I will try that thank u
Imani
Case of Equilateral Hyperbola
Jhon Reply
ok
Zander
ok
Shella
f(x)=4x+2, find f(3)
Benetta
f(3)=4(3)+2 f(3)=14
lamoussa
14
Vedant
pre calc teacher: "Plug in Plug in...smell's good" f(x)=14
Devante
8x=40
Chris
Explain why log a x is not defined for a < 0
Baptiste Reply
the sum of any two linear polynomial is what
Esther Reply
divide simplify each answer 3/2÷5/4
Momo Reply
divide simplify each answer 25/3÷5/12
Momo
how can are find the domain and range of a relations
austin Reply
the range is twice of the natural number which is the domain
Morolake
A cell phone company offers two plans for minutes. Plan A: $15 per month and $2 for every 300 texts. Plan B: $25 per month and $0.50 for every 100 texts. How many texts would you need to send per month for plan B to save you money?
Diddy Reply
6000
Robert
more than 6000
Robert
For Plan A to reach $27/month to surpass Plan B's $26.50 monthly payment, you'll need 3,000 texts which will cost an additional $10.00. So, for the amount of texts you need to send would need to range between 1-100 texts for the 100th increment, times that by 3 for the additional amount of texts...
Gilbert
...for one text payment for 300 for Plan A. So, that means Plan A; in my opinion is for people with text messaging abilities that their fingers burn the monitor for the cell phone. While Plan B would be for loners that doesn't need their fingers to due the talking; but those texts mean more then...
Gilbert
can I see the picture
Zairen Reply
How would you find if a radical function is one to one?
Peighton Reply
how to understand calculus?
Jenica Reply
with doing calculus
SLIMANE
Thanks po.
Jenica
Hey I am new to precalculus, and wanted clarification please on what sine is as I am floored by the terms in this app? I don't mean to sound stupid but I have only completed up to college algebra.
rachel Reply
I don't know if you are looking for a deeper answer or not, but the sine of an angle in a right triangle is the length of the opposite side to the angle in question divided by the length of the hypotenuse of said triangle.
Marco
can you give me sir tips to quickly understand precalculus. Im new too in that topic. Thanks
Jenica
if you remember sine, cosine, and tangent from geometry, all the relationships are the same but they use x y and r instead (x is adjacent, y is opposite, and r is hypotenuse).
Natalie
it is better to use unit circle than triangle .triangle is only used for acute angles but you can begin with. Download any application named"unit circle" you find in it all you need. unit circle is a circle centred at origine (0;0) with radius r= 1.
SLIMANE
What is domain
johnphilip
the standard equation of the ellipse that has vertices (0,-4)&(0,4) and foci (0, -15)&(0,15) it's standard equation is x^2 + y^2/16 =1 tell my why is it only x^2? why is there no a^2?
Reena Reply
Practice Key Terms 4

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Source:  OpenStax, Precalculus. OpenStax CNX. Jan 19, 2016 Download for free at https://legacy.cnx.org/content/col11667/1.6
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