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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
Is the product of two imaginary numbers always an imaginary number?
If you are talking about pure imaginary numbers (a complex number with no real part) then no. Example: bi times ci where b and c are real numbers equals b*c*i² = b*c*(-1) = -b*c, which is a real number, because b & c & -1 are all real numbers. If you're talking about multiplying two complex numbers (a + bi)*(c + di), then the product will be complex, but it could be real or imaginary, depending on the values of a, b, c, & d.
Why are imaginary numbers important?
They are used for working out equations where the numbers you are working with are not physically possible, but we just imagine they are, such as the square root of a negative number
In engineering, especially Electrical Engineering, using complex numbers to represent signals (rather than sines and/or cosines) make comparing and working with signals easier.
What is the Multiplicative inverse formula of complex numbers?
So if you have a number z = a + bi. Then how to find 1 divided by z. The way to figure this is to get the denominator as a pure real number. Multiplying the numerator and the denominator by the complex conjugate {a - bi} will result in a pure real denominator.
(a - bi)(a + bi) = a² + abi - abi - (bi)² = a² + b². So the multiplicative inverse is
(a - bi)/(a² + b²)
Is the product of a complex number and its conjugate a real number?
Yes. If you multiply X + iY by X - iY you get X2 + Y2. The imaginary parts cancel out.
A complex number might not be a pure imaginary number?
True. Complex numbers have a real part and an imaginary part. If either one of these is zero, the complex number will be a pure real or a pure imaginary.
What is the magnitude of the complex number 3-9i?
Mag(3 - 9i) = sqrt(32 + 92) = sqrt(9 + 81) = sqrt(90) or 3*sqrt(10)
What is the definition of a pure imaginary number?
A pure imaginary number is a complex number that has 0 for its real part, such as 0+7i.
What is the square root of negative 54 using imaginary numbers?
±3i√6
Rounded to two decimal places, the square root of +54 is equal to ±7.35. Therefore, the square root of -54, rounded to two decimal places, is equal to ±7.35 i.
Are skew symmetric roots purely real or purely imaginary?
They can be either. If they are roots of a real polynomial then purely imaginary would be symmetric and only real roots can be skew symmetric.
What is an example of rote counting?
Teacher"Show me how you counted to ten" Student "Like this:two,four,six"
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.
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.
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
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.
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 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.
