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Complex Numbers
The square root of negative one, which we now call the imaginary unit, was once thought to be an absurd notion. However, through the diligent studies of open-minded mathematicians, it was shown that the real numbers were actually just one part of a larger set of numbers known as the complex numbers, the other part being the imaginary numbers. Please direct all questions about these surprisingly useful and applicable numbers into this category.
887 Questions
You cannot show it in general since it need not be true!
Applications and use of complex numbers in engineering?
Any physical motion which is periodic, such as an oscillating beam, string, wire, pendulum, electronic signal, or electromagnetic wave can be represented by a complex number function. This can make calculations with the various components simpler than with real numbers and sines and cosines.
How the set of real no is uncountable?
There is no one to one correspondence between the real numbers and the set of integers. In fact, the cardinality of the real numbers is the same as the cardinality of the power set of the set of integers, that is, the set of all subsets of the set of integers.
What are the applications of complex numbers in real life?
- As with many other topics of advanced math, it depends whether your "real life" includes working in some engineering area, or not. If you work in electrical engineering or electronics, you will use complex numbers on a daily basis; similarly if you do a lot of math for some reason.
- Complex Numbers (ones involving the imaginary unit 'i' which is defined to handle the square root of -1) are a convenient way to describe the behavior of some electric circuits and waves.
- One thing you may encounter in your 'real life' are uninterruptible power supplies (UPS - what keeps your computer running when the power goes out). You will see ratings such as 500 VA / 300 W. This has to do with Complex Power. Electric circuits, which have inductances (such as transformers and electric motors) or capacitors, behave in a way that is described by complex numbers. While Volts X Amps equals Watts, the 300 W is the 'real power' (the amount available to do work), and the 500 VA is the apparent power (the total power that the power supply is capable of producing). The reactive power (due to capacitance & inductance) is represented by the imaginary component and it is at right angles to the real power (represented by the real component).
- Another use of complex numbers is in the Fast Fourier Transform (FFT) - one of the most ubiquitous algorithms - used heavily for signal processing. If you have a digital camera, a cell phone, an LCD - FFT is there, bringing complex numbers along.
- The complex equation w=z+1/z is a basic formula used for designing air foils-airplane wings and Figuring out flow forces around a circular object in water for instance.
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Are imaginary numbers rational or irrational?
Imaginary numbers are not intrinsically rational or irrational.
Of course, all real numbers are either rational or irrational numbers.
Imaginary numbers are not real numbers.
Imaginary numbers have a real part and an imaginary part, sometimes written like z=x+i y.
The two parts, i.e. the x and the y, are real numbers. As real numbers, they are either rational or irrational. Its just that the two parts of a complex number may both be either rational or irrational or one may be rational and the other irrational. One could always make up a new name for these cases, but right now there is no such classification.
How do you create a complex formula that calculates a 30 percent decrease?
We're stumped. The only formula up with which we're able to come is a simple one.
If the original quantity is 'Q', then after a 30-percent decrease, the new quantity 'N' is:
N = 0.7 Q
What is the relationship between a complex number and its conjugate?
When a complex number is multiplied by its conjugate, the product is a real number and the imaginary number disappears.
Why must you rationalize complex numbers and not divide them?
Since a Complex number has a real component and an imaginary component, it would be like trying to divide z / (2x + 3y)
What is an example of rote counting?
Teacher"Show me how you counted to ten" Student "Like this:two,four,six"
Do all real numbers lie on a single line in the complex plane?
Yes. Traditionally, this line is drawn horizontally, with positive numbers to the right, and negative numbers to the left.
The set of complex numbers is the set of numbers which can be described by a + bi, where a and b are real numbers, and i is the imaginary unit sqrt(-1). Since a and b can be any real number (including zero), the set of real numbers is a subset of the set of complex numbers. Also the set of pure imaginary numbers is a subset of complex number set.
Which number belongs to the set of complex numbers?
Lots of numbers do.
To begin, all real numbers do.
Multiples of sqrt(-1), aka. imaginary numbers, do.
The Complex Numbers are all numbers which are the sum of a real number and an imaginary number.
How do you find cube roots of imaginary numbers?
For a pure imaginary number: {i = sqrt(-1)} times a real coefficient {r}, you have i*r. The cube_root(i*r) = cube_root(i)*cube_root(r), so find the cube root of r in the normal way, then we just need to find the cube root of i. For any cubic function (which has a polynomial, in which the highest term is x3) will always have 3 roots. There are 3 values, which when cubed will equal the imaginary number i:
- -i will do it: (-i)3 = (-i)2 * (-i) = -1 * (-i) = i
- The other two are complex: sqrt(3)/2 + i/2 and -sqrt(3)/2 + i/2
If you cube either of the two complex binomials by multiplying out, you will end up with 0 + i as the answer in both cases.
Note: the possible roots for any cubic are: 3 real roots, or 1 real root and 2 complex root, or 1 pure imaginary root, and 2 complex roots.
For your original question, if you want to stay in the pure imaginary domain, then you can use: Cube_root(i*r) = -i * cube_root(r) to find an answer.
This forms an equation which solves nicely:
2x-7 = x-1
x-7 = -1
x = 6
What is seven eighths times two fifths?
When multiplying fractions we multiply all the numerators to produce a new numerator and we multiply all the denominators to produce a new denominator, then we reduce the new fraction to its simplest form.
In this case - 7/8 x 2/5 gives 7 x 2 = 14 as the new numerator and 8 x 5 = 40 as the new denominator
The "new" fraction is now 14/40 or 7/20
What are the 3 complex roots of -1 using the DeMoivre's theorem?
Problem: find three solutions to z^3=-1.
DeMoivre's theorem is that (cos b + i sin b)^n = cos bn + i sin bn
So we can set
z= (cos b + i sin b),
n = 3
cos bn + i sin bn = -1.
From the last equation, we know that cos bn = -1, and sin bn = 0.
Three possible solutions are bn=pi, bn=3pi, bn=5pi. This gives three possible values of b:
b=pi/3
b=pi
b = 5pi/3.
Now using z= (cos b + i sin b), we can get three possible cube roots of -1:
z= (cos pi/3 + i sin pi/3),
z= (cos pi + i sin pi),
z= (cos 5pi/3 + i sin 5pi/3).
Working these out gives
-1/2+i*sqrt(3)/2
-1
-1/2-i*sqrt(3)/2
Is two thirds and four ninths equivalent?
Umm, no. Two thirds and six ninths are equivalent. They look like this 2/3 and 6/9, to get to 9 you times the 3 by 3 so to get the 6 you have to times the 2 by 3. I can see that you've seen that you have squared 3 so you had to square 2, but unfortunately that isn't what happens.
