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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.
8 less than a number?
8 less than a number would be written as 8 - 5. or 8 - A No! 8 less than a number, say X, would be X - 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 complex conjugate of 8?
8
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 be 8 but adds to be negative 4?
12
What multiplies to 105 and adds to 8?
113
