The answer depends on the context, but bi is a variable, not a number. It could be the ith of a set of variables b1, b2, ... .
Or it could be the square root of -b2.
Or it could be a vector of magnitude b in the idirection.
A number of the form (a + bi) is a complex number.
Complex
complex
"a + bi" is a common way to write a complex number. Here, "a" and "b" are real numbers.Another common way to write a complex number is in polar coordinates - basically specifying the distance from zero, and an angle.
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)
Given a complex number z = a + bi, the conjugate z* = a - bi, so z + z*= a + bi + a - bi = 2*a. Note that a and b are both real numbers, and i is the imaginary unit: +sqrt(-1).
The prefix Bi- represents the number 2.
Yes. By definition, the complex conjugate of a+bi is a-bi and a+bi - (a - bi)= 2bi which is imaginary (or 0)
Complex numbers form: a + bi where a and b are real numbers. The conjugate of a + bi is a - bi If you multiply a complex number by its conjugate, the product will be a real number, such as (a + bi)(a - bi) = a2 - (bi)2 = a2 - b2i2 = a2 - b2(-1) = a2 + b2
A number of the form (a + bi) is a complex number.
2
Complex
Yes, the difference between a complex number and its conjugate is a pure imaginary number. If we represent a complex number as ( z = a + bi ) (where ( a ) is the real part and ( b ) is the imaginary part), its conjugate is ( \overline{z} = a - bi ). The difference ( z - \overline{z} = (a + bi) - (a - bi) = 2bi ), which is purely imaginary since it has no real part.
Bi shows the element bismuth. Atomic number if Bi is 83.
The oxidation number of Bi in Bi(OH)3 is +3. Oxygen generally has an oxidation number of -2, and since there are three hydroxide (OH-) ions, each with a -1 charge, the oxidation number of Bi can be calculated as follows: x + 3(-2) + 3(-1) = 0. Solve for x to get x = +3.
It is a pure imaginary number.Since (a+bi)-(a-bi) = 2bi, it is a pure imaginary number (it has no real component).
The phone number of the Bloomer Bi-Centennial House is: 715-568-1776.