0
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
What is a leprecon is he real or imaginary?
Leprecon's only exist after generous portions of Irish stout. How do you know if it's a "he"?
How do you simplify 9x plus 6-4x-2x plus 1-15?
Assuming the equation is exactly as you described in words, it would be expressed algebraically as 9X+(6-4X-2X)+(1-15)
9X+(6-4X-2X)+(1-15) First, Combine all like terms inside the parenthesis
9X+(6-6X)+(-14) Next, clear the parenthesis. This is done by distributing the "1" coefficient implied by the parenthesis over the entire parenthesized term.
9X+1(6-6X)+1(-14)
9X+6-6X-14 Combine like terms again to get your answer!
=3X-8
What is the significance of complex roots?
One significant feature of complex numbers is that all polynomial equations of order n, in the complex field, have n solutions. When multiple roots are
Given any set of complex numbers {a(0),  … , a(n)}, such that at least one of a(1) to a(n) is non-zero, the equation
a(n)*z^n + a(n-1)*z^(n-1) + ... + a(0) has at least one solution in the complex field.
This is the Fundamental Theorem of Algebra and establishes the set of Complex numbers as a closed field.
[a(0), ... , a(n) should be written with suffices but this browser has decided not to be cooperative!]
The above solution is the complex root of the equation.
In fact, if the equation is of order n, that is, if the coefficient a(n) is non-zero then, taking account of the multiplicity, the equation has exactly n roots (some of which may be real).
How do you simplify complex numbers?
Hi,
See the related answers and the Related Link. My friends are working hard to supply
more answers.
What is the square root of -45 using imaginary numbers?
Square root of -45 = (-45)1/2 = (-3x3x5)1/2 = 3(-5)1/2 = 351/2i
What set of numbers does 30 belong to?
To any set that contains it!
It belongs to {30},
or {45, sqrt(2), 30, pi, -3/7},
or all whole numbers between 23 and 53,
or multiples of 5,
or square roots of 900,
or composite numbers,
or counting numbers,
or integers,
or rational numbers,
or real numbers,
or complex numbers,
etc.
How do you find the additive inverse of a complex number?
You take the additive invers of the real and of the imaginary part. For instance, the additive inverse of: (3 - 5i) is (-3 + 5i).You take the additive invers of the real and of the imaginary part. For instance, the additive inverse of: (3 - 5i) is (-3 + 5i).
You take the additive invers of the real and of the imaginary part. For instance, the additive inverse of: (3 - 5i) is (-3 + 5i).You take the additive invers of the real and of the imaginary part. For instance, the additive inverse of: (3 - 5i) is (-3 + 5i).
You take the additive invers of the real and of the imaginary part. For instance, the additive inverse of: (3 - 5i) is (-3 + 5i).You take the additive invers of the real and of the imaginary part. For instance, the additive inverse of: (3 - 5i) is (-3 + 5i).
You take the additive invers of the real and of the imaginary part. For instance, the additive inverse of: (3 - 5i) is (-3 + 5i).You take the additive invers of the real and of the imaginary part. For instance, the additive inverse of: (3 - 5i) is (-3 + 5i).
complex to-complex(mag, theta)
{
if(mag >=0)
{
real = mag * cos(theta);
img = mag * sin(theta);
return real + i * img;
}
else
raise error;
}
One test is needed : mag must be a positive number!
And the return value is depending of your way to deal with complex number
Plot -5 plus 9i in a complex plane?
Start where the x and y axes cross.
Go 5 units to the left on the x (horizontal axis) and then go 9 units up from there.
Put a dot in that spot.
If the discriminant is zero then there are no imaginary solutions?
Yes, if the discriminant is zero, then there will be a double root, which will be real.
Also, If the discriminant is positive, there will be two distinct real solutions. But if the discriminant is negative, then you will have two complex solutions.
Can x to the fourth power ever result in a negative number?
no even exponent of a real number can ever result in a negative number. If x is a complex number with the real and imaginary part having the same magnitude, then taking that to the fourth power will result in a real number, which is negative.
Example: (2 + 2i)4, or (-2 + 2i)4, or (2 - 2i)4, or (-2 - 2i)4, Just take (2 - 2i)4, as one to see how it works. First take (2 - 2i)2, then we'll square that result.
(2 - 2i)2 = 4 - 4i - 4i + 4i2 , but i2 is -1, so we have -8i, then square that is 64i2 which is -64.
Is the product of two conjugate complex number always a real number?
Yes. This can be verified by using a "generic" complex number, and multiplying it by its conjugate:
(a + bi)(a - bi) = a2 -abi + abi + b2i2 = a2 - b2
Alternative proof:
I'm going to use the * notation for complex conjugate. Any complex number w is real if and only if w=w*. Let z be a complex number. Let w = zz*. We want to prove that w*=w. This is what we get:
w* = (zz*)*
= z*z** (for any u and v, (uv)* = u* v*)
= z*z
= w
Find the complex fourth root of 256i Express your answer in a plus bi form?
The four roots of 4√256 are {4, -4, 4i, and -4i}. Note that two of them are real numbers and the other two are pure imaginary, therefore 0 + 4i is the same as just 4i
What is the name for 1000000000000000000000000000000000000000000000000000000000000000000?
For such large numbers, just write the number in scientific notation, like this: 10x, replacing "x" with the number of zeroes.
What are the engineering applications of complex numbers and matrices?
I suggest asking separate questions for complex numbers, and for matrices.
Complex numbers are used in a variety of fields, one of them is electrical engineering. As soon as AC circuits are analyzed, it turns out that complex numbers are the natural way to do this.
How do you calculate root of any complex number using casio Fx 991MS calculator?
Not sure about the Casio, but most calculators which have capability to handle complex numbers should be similar. Input the complex number according to however you normally do that, then raise to a power. In the case of roots, you want to raise to a reciprocal power: Square root is 0.5 power, cube root is 1/3 power, fourth root is 0.25 power, etc
Does trichotomy axiom hold true for complex numbers?
I think so. In x+iy, x and y are real numbers and have to be <0,0 or >0.
How do you enter an imaginary number in a scientific calculator?
Some scientific calculators can't handle complex or imaginary numbers. If you happen to have a special calculator that does, probably the manual will tell you how to enter them.
The HP 48 and up series does. It depends on if your calculator is in Polar Coordinate mode or X-Y coordinate mode, but a quick way to get the imaginary number i (regardless of which mode the calculator is currently in), is to press -1, then 'square root' button.
A non complex number is a number that does not have any imaginary component. An imaginary component is a non zero factor of the square root of -1, in other words, the imaginary number i.
Square root of 17 in complex number terms?
A positive real number, such as 17, has two square roots. One is the one your calculator gives you, if you use the square root function. The other is the same number, with a minus sign in front. None of these has an imaginary part. There are no additional complex roots that have a non-zero imaginary part.
