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This module introduces the concept of complex numbers in Algebra.

A “complex number” is the sum of two parts: a real number by itself, and a real number multiplied by i . It can therefore be written as a + bi , where a and b are real numbers.

The first part, a , is referred to as the real part . The second part, bi , is referred to as the imaginary part .

Examples of complex numbers a + bi ( a is the “real part”; bi is the “imaginary part”)
3 + 2i a = 3 , b = 2
π a = π , b = 0 (no imaginary part: a “pure real number”)
-i a = 0 , b = -1 (no real part: a “pure imaginary number”)

Some numbers are not obviously in the form a + bi . However, any number can be put in this form.

Example 1: Putting a fraction into a + bi form ( i in the numerator)
3 4i 5 size 12{ { {3 - 4i} over {5} } } {} is a valid complex number. But it is not in the form a + bi , and we cannot immediately see what the real and imaginary parts are.
To see the parts, we rewrite it like this:
3 4i 5 size 12{ { {3 - 4i} over {5} } } {} = 3 5 size 12{ { {3} over {5} } } {} 4 5 size 12{ { {4} over {5} } } {} i
Why does that work? It’s just the ordinary rules of fractions, applied backward. (Try multiplying and then subtracting on the right to confirm this.) But now we have a form we can use:
3 4i 5 size 12{ { {3 - 4i} over {5} } } {} a = 3 5 size 12{ { {3} over {5} } } {} , b = 4 5 size 12{ { {4} over {5} } } {}
So we see that fractions are very easy to break up, if the i is in the numerator. An i in the denominator is a bit trickier to deal with.
Example 2: Putting a fraction into a + bi form ( i in the denominator)
1 i size 12{ { {1} over {i} } } {} = 1 i i i size 12{ { {1 cdot i} over {i cdot i} } } {} Multiplying the top and bottom of a fraction by the same number never changes the value of the fraction: it just rewrites it in a different form.
= i 1 size 12{ { {i} over { - 1} } } {} Because i i is i 2 , or –1.
= -i This is not a property of i , but of –1. Similarly, 5 1 size 12{ { {5} over { - 1} } } {} = –5 .
1 i size 12{ { {1} over {i} } } {} : a = 0 , b = -1 since we rewrote it as -i , or 0 - 1i

Finally, what if the denominator is a more complicated complex number? The trick in this case is similar to the trick we used for rationalizing the denominator: we multiply by a quantity known as the complex conjugate of the denominator .

Definition of complex conjugate

The complex conjugate of the number a + bi is a - bi . In words, you leave the real part alone, and change the sign of the imaginary part.

Here is how we can use the “complex conjugate” to simplify a fraction.

Example: Using the Complex Conjugate to put a fraction into a + bi form
5 3 4i size 12{ { {5} over {3 - 4i} } } {} The fraction: a complex number not currently in the form a + b i
= 5 ( 3 + 4i ) ( 3 4i ) ( 3 + 4i ) size 12{ { {5` \( 3+4i \) } over { \( 3 - 4i \) \( 3+4i \) } } } {} Multiply the top and bottom by the complex conjugate of the denominator
= 15 + 20 i 3 2 ( 4i ) 2 size 12{ { {"15"+"20"i} over {3 rSup { size 8{2} } - \( 4i \) rSup { size 8{2} } } } } {} Remember, ( x + a ) ( x a ) = x 2 –a 2
= 15 + 20 i 9 + 16 size 12{ { {"15"+"20"i} over {9+"16"} } } {} ( 4i ) 2 = 4 2 i 2 = 16 ( –1 ) = –16 , which we are subtracting from 9
= 15 + 20 i 25 size 12{ { {"15"+"20"i} over {"25"} } } {} Success! The top has i , but the bottom doesn’t. This is easy to deal with.
= 15 25 size 12{ { {"15"} over {"25"} } } {} + 20 i 25 size 12{ { {"20"i} over {"25"} } } {} Break the fraction up, just as we did in a previous example.
= 3 5 size 12{ { {3} over {5} } } {} + 4 5 size 12{ { {4} over {5} } } {} i So we’re there! a = 3 5 size 12{ { {3} over {5} } } {} and b = 4 5 size 12{ { {4} over {5} } } {}

Any number of any kind can be written as a + bi . The above examples show how to rewrite fractions in this form. In the text, you go through a worksheet designed to rewrite 1 3 size 12{ nroot { size 8{3} } { - 1} } {} as three different complex numbers. Once you understand this exercise, you can rewrite other radicals, such as i size 12{ sqrt {i} } {} , in a + bi form.

Questions & Answers

can someone help me with some logarithmic and exponential equations.
Jeffrey Reply
sure. what is your question?
ninjadapaul
20/(×-6^2)
Salomon
okay, so you have 6 raised to the power of 2. what is that part of your answer
ninjadapaul
I don't understand what the A with approx sign and the boxed x mean
ninjadapaul
it think it's written 20/(X-6)^2 so it's 20 divided by X-6 squared
Salomon
I'm not sure why it wrote it the other way
Salomon
I got X =-6
Salomon
ok. so take the square root of both sides, now you have plus or minus the square root of 20= x-6
ninjadapaul
oops. ignore that.
ninjadapaul
so you not have an equal sign anywhere in the original equation?
ninjadapaul
Commplementary angles
Idrissa Reply
hello
Sherica
im all ears I need to learn
Sherica
right! what he said ⤴⤴⤴
Tamia
what is a good calculator for all algebra; would a Casio fx 260 work with all algebra equations? please name the cheapest, thanks.
Kevin Reply
a perfect square v²+2v+_
Dearan Reply
kkk nice
Abdirahman Reply
algebra 2 Inequalities:If equation 2 = 0 it is an open set?
Kim Reply
or infinite solutions?
Kim
The answer is neither. The function, 2 = 0 cannot exist. Hence, the function is undefined.
Al
y=10×
Embra Reply
if |A| not equal to 0 and order of A is n prove that adj (adj A = |A|
Nancy Reply
rolling four fair dice and getting an even number an all four dice
ramon Reply
Kristine 2*2*2=8
Bridget Reply
Differences Between Laspeyres and Paasche Indices
Emedobi Reply
No. 7x -4y is simplified from 4x + (3y + 3x) -7y
Mary Reply
is it 3×y ?
Joan Reply
J, combine like terms 7x-4y
Bridget Reply
im not good at math so would this help me
Rachael Reply
yes
Asali
I'm not good at math so would you help me
Samantha
what is the problem that i will help you to self with?
Asali
how do you translate this in Algebraic Expressions
linda Reply
Need to simplify the expresin. 3/7 (x+y)-1/7 (x-1)=
Crystal Reply
. After 3 months on a diet, Lisa had lost 12% of her original weight. She lost 21 pounds. What was Lisa's original weight?
Chris Reply
what's the easiest and fastest way to the synthesize AgNP?
Damian Reply
China
Cied
types of nano material
abeetha Reply
I start with an easy one. carbon nanotubes woven into a long filament like a string
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many many of nanotubes
Porter
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Yasmin
what is the function of carbon nanotubes?
Cesar
what is nanomaterials​ and their applications of sensors.
Ramkumar Reply
what is nano technology
Sravani Reply
what is system testing?
AMJAD
preparation of nanomaterial
Victor Reply
Yes, Nanotechnology has a very fast field of applications and their is always something new to do with it...
Himanshu Reply
good afternoon madam
AMJAD
what is system testing
AMJAD
what is the application of nanotechnology?
Stotaw
In this morden time nanotechnology used in many field . 1-Electronics-manufacturad IC ,RAM,MRAM,solar panel etc 2-Helth and Medical-Nanomedicine,Drug Dilivery for cancer treatment etc 3- Atomobile -MEMS, Coating on car etc. and may other field for details you can check at Google
Azam
anybody can imagine what will be happen after 100 years from now in nano tech world
Prasenjit
after 100 year this will be not nanotechnology maybe this technology name will be change . maybe aftet 100 year . we work on electron lable practically about its properties and behaviour by the different instruments
Azam
name doesn't matter , whatever it will be change... I'm taking about effect on circumstances of the microscopic world
Prasenjit
how hard could it be to apply nanotechnology against viral infections such HIV or Ebola?
Damian
silver nanoparticles could handle the job?
Damian
not now but maybe in future only AgNP maybe any other nanomaterials
Azam
can nanotechnology change the direction of the face of the world
Prasenjit Reply
At high concentrations (>0.01 M), the relation between absorptivity coefficient and absorbance is no longer linear. This is due to the electrostatic interactions between the quantum dots in close proximity. If the concentration of the solution is high, another effect that is seen is the scattering of light from the large number of quantum dots. This assumption only works at low concentrations of the analyte. Presence of stray light.
Ali Reply
the Beer law works very well for dilute solutions but fails for very high concentrations. why?
bamidele Reply
how did you get the value of 2000N.What calculations are needed to arrive at it
Smarajit Reply
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Source:  OpenStax, Math 1508 (lecture) readings in precalculus. OpenStax CNX. Aug 24, 2011 Download for free at http://cnx.org/content/col11354/1.1
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