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Is the difference of a complex number and it's conjugate an imaginary number?
Yes. By definition, the complex conjugate of a+bi is a-bi and a+bi - (a - bi)= 2bi which is imaginary (or 0)
What is the complex conjugate of 8 plus 9i?
It would be 8 minus 9i or 8-9i
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
What is the complex conjugate of 2-3i?
The complex conjugate of 2-3i is 2+3i.
What is the multiplicative inverse of 4 plus i?
The multiplicative inverse of a complex number is the reciprocal of that number. To find the multiplicative inverse of 4 + i, we first need to find the conjugate of 4 + i, which is 4 - i. The product of a complex number and its conjugate is always a real number. Therefore, the multiplicative inverse of 4 + i is (4 - i) / ((4 + i)(4 - i)) = (4 - i) / (16 + 1) = (4 - i) / 17.
If a and b are any real numbers what is the conjugate of a plus b?
The concept of conjugate is usually used in complex numbers. If your complex number is a + bi, then its conjugate is a - bi.
What is the complex conjugate of a-bi?
a+bi
How do you simplify a complex fraction?
You multiply the numerator and the denominator of the complex fraction by the complex conjugate of the denominator.The complex conjugate of a + bi is a - bi.
What is the product of the complex number a plus bi and its conjugate?
The product is a^2 + b^2.
How do you simplify complexed fractions?
You multiply the numerator and the denominator of the complex fraction by the complex conjugate of the denominator.The complex conjugate of a + bi is a - bi.
Is the difference of a complex number and it's conjugate an imaginary number?
Yes. By definition, the complex conjugate of a+bi is a-bi and a+bi - (a - bi)= 2bi which is imaginary (or 0)
How do conjugate arrive at complex number?
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
What is the complex conjugate of a plus bi?
The complex conjugate of a+bi is a-bi. This is written as z* where z is a complex number. ex. z = a+bi z* = a-bi r = 3+12i r* = 3-12i s = 5-6i s* = 5+6i t = -3+7i = 7i-3 t* = -3-7i = -(3+7i)
What is the meaning of Complex conjugate reflection?
For a complex number (a + bi), its conjugate is (a - bi). If the number is graphically plotted on the Complex Plane as [a,b], where the Real number is the horizontal component and Imaginary is vertical component, the Complex Conjugate is the point which is reflected across the real (horizontal) axis.
The sum of a complex number and its conjugate?
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).
Why do you multiply by the complex conjugate?
Whenever a complex number (a + bi) is multiplied by it's conjugate (a - bi), the result is a real number: (a + bi)* (a - bi) = a2 - abi + abi - (bi)2 = a2 - b2i2 = a2 - b2(-1) = a2 + b2 This is useful when dividing complex numbers, because the numerator and denominator can both be multiplied by the denominator's conjugate, to give an equivalent fraction with a real-number denominator.
What is the complex conjugate of the following complex number 7 plus 5i?
The conjugate is 7-5i
