(3-4i)(1-i) = (3x1) + (3 x -i) + (-4i x 1) + ( -4i x -i) = 3 - 3i -4i -4 = -1 - 7i
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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.