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What is the complex conjugate complex number-8-6i?
-6i-8
The sum of two complex numbers is always a complex number?
A "complex number" is a number of the form a+bi, where a and b are both real numbers and i is the principal square root of -1. Since b can be equal to 0, you see that the real numbers are a subset of the complex numbers. Similarly, since a can be zero, the imaginary numbers are a subset of the complex numbers. So let's take two complex numbers: a+bi and c+di (where a, b, c, and d are real). We add them together and we get: (a+c) + (b+d)i The sum of two real numbers is always real, so a+c is a real number and b+d is a real number, so the sum of two complex numbers is a complex number. What you may really be wondering is whether the sum of two non-real complex numbers can ever be a real number. The answer is yes: (3+2i) + (5-2i) = 8. In fact, the complex numbers form an algebraic field. The sum, difference, product, and quotient of any two complex numbers (except division by 0) is a complex number (keeping in mind the special case that both real and imaginary numbers are a subset of the complex numbers).
An acid has Ka equals 4.5 x 10- 7 What is the Kb for the conjugate base of this acid at 25 C?
2.2 x 10-8
Is 8 is an real number?
Nearly any number you can think of is a Real Number. So 8 is a real number.
A number is divisible by 8 the number formed by its last?
If the last three digits of a number are divisible by 8, the whole number is divisible by 8.
What is the complex conjugate complex number-8-6i?
-6i-8
What is the conjugate of -8-4i?
The conjugate of -8-4i is -8+4i. It is obtained by changing the sign of the imaginary part of the complex number.
What is the conjugate of 8-3i?
11
What is the complex conjugate of 8?
Oh, isn't that just a happy little question? The complex conjugate of a real number like 8 is just 8 itself because there is no imaginary part to change. Just like how every tree needs its roots, every real number needs its complex conjugate to stay balanced and harmonious. Just remember, there are no mistakes, only happy little accidents in math!
What is the complex conjugate of 8 plus 8i?
8 - 8i
What is the complex form of 4 plus 4i plus 4 plus 6i?
the problem: what is 4 + 4i + 4 + 6i what you do is add the real and imaginary parts, thus: 4+4 and 4i+6i = 8+10i answer.
What is the complex conjugate of 8 plus 9i?
It would be 8 minus 9i or 8-9i
8 plus 6i-2 plus 3i?
(8+6i)-(2+3i)=6+3i 8+6i-2+3i=6+9i
What is The square root of 25-2times the square root of -36 plus the square root of four - the square root of 121?
Square root of 25 = +or- 5 Square root of -36 = +or- 6i where i is the imaginary number such that i^2=-1 Square root of 121 = +or-11 So the 8 possible answers are: -16-6i, -16+6i, -6-6i, -6+6i, 6-6i, 6+6i, 16-6i and 16+6i
What number add to give 8 but multiply to give you 154?
162
How do you divide complex numbers?
When dividing complex numbers you must:Write the problem in fractional formRationalize the denominator by multiplying the numerator and denominator by the conjugate of the denominator.You must remember that a complex number times its conjugate will give a real number.a complex number 2+2i. the conjugate to this is 2-i1. Multiply both together gives a real number.(2+2i)(2-2i) = 4 -4i + 4i + (-4i2) (and as i2 = -1) = 8To divide a complex number by a real number simply divide the real parts by the divisor.(8+4i)/2 = (4+2i)To divide a real number by a complex number.1. make a fraction of the expression 8/(2+2i)2. multiply by 1. express 1 as a fraction of the divisor's conjunction. 8/(2+2i)*(2-2i)/(2-2i)3. multiply numerator by numerator and denominator by denominator.(16-16i)/84. and simplify 2-2i
What multiplies to 105 and adds to 8?
113
