Complex math covers how to do operations on complex numbers. Complex numbers include real numbers, imaginary numbers, and the combination of real+imaginary numbers.
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).
If z1=a+ib and z2=c+id then the product z1*z2=(ac-bd)+i(ad+bc)
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.
The complex numbers are a field.
No, they cannot with real numbers. With complex numbers it is possible.
If you add two complex numbers, the resulting complex number is equivalent to the vector resulting from adding the two vectors. If you multiply two complex numbers, the resulting complex number is equivalent to the vector resulting from the cross product of the two vectors.
Complex numbers, Real numbers, Rational numbers, Integers, Natural Numbers, Multiples of an integer.
Real numbers are a proper subset of complex numbers. In fact each complex number, z, can be represented as z = x +iy where x and y are real numbers and i is the imaginary square root of -1.Thus the set of complex numbers is the Cartesian product of two sets of real numbers. That is, C = R x R where C is the set of complex numbers and R is the set of real numbers. Limitations of this browser prevent me from writing that in a mathematically precise and more helpful fashion.
The question has no answer in real numbers. The solution, in complex numbers, are 2+3i and 2-3i where i is the imaginary square root of -1.
The two numbers are the complex conjugate pair27.5 - 79.0174iand27.5 + 79.0174iwhere i is the imaginary square root of -1.
The two numbers are the complex conjugate pair, 3.5 - 7.6i and 3.5 + 7.6i where i is the imaginary root of -1
There are no real numbers that satisfy the requirements. The complex solutions are: 1 +/- sqrt(21)
That is the identity property of multiplication for all rational numbers, or all real numbers or all complex numbers except (in each case) for 0.
Complex math covers how to do operations on complex numbers. Complex numbers include real numbers, imaginary numbers, and the combination of real+imaginary numbers.
Complex numbers are a proper superset of real numbers. That is to say, real numbers are a proper subset of complex numbers.
No. Complex numbers is the highest set of numbers you can go, and there are no sets outside of complex numbers.