Add the real and the imaginary parts separately.
-6-4i.
The additive inverse of 6+4i is -6-4i since their sum is 0. It is analogous to real numbers where the additive inverse of 6 is -6 since 6+-6 =6-6=0 In the case of complex numbers, we add them by adding the real parts and then adding the imaginary parts. So to find the complex additive inverse of a+bi, we find the inverse of a which is -a and of bi which is -bi and so the additive inverse is -a-bi
The sum of -8 and 6 is -2
6
A product is a binary operatoin. That is, it requires two numbers to be combined. There is only one number, 2 + 4i, in the question.
(2 + 4i) - (7 + 4i) = -5 2 + 4i - 7 + 4i = -5 + 8i
-6-4i.
Since the imaginary parts cancel, and the real parts are the same, the sum is twice the real part of any of the numbers. For example, (5 + 4i) + (5 - 4i) = 5 + 5 + 4i - 4i = 10.
To find the sum of (12 + 5i) and (3 + 4i), you add the real parts and the imaginary parts separately. The real part is (12 + 3 = 15) and the imaginary part is (5i + 4i = 9i). Therefore, the sum is (15 + 9i).
The additive inverse of 6+4i is -6-4i since their sum is 0. It is analogous to real numbers where the additive inverse of 6 is -6 since 6+-6 =6-6=0 In the case of complex numbers, we add them by adding the real parts and then adding the imaginary parts. So to find the complex additive inverse of a+bi, we find the inverse of a which is -a and of bi which is -bi and so the additive inverse is -a-bi
4i(-2 -3i) = 4i×-2 - 4i×-3i = -8i -12i² = -8i + 12 = 12 -8i → the conjugate is 12 + 8i
(9 + 4i)^2 = 9^2 + (2)(9)(4i) +i^2 substitute i^2 for -1; = 81 + 72i -1 = 80 + 72i
the sum of -2 and 6 and 7 is (-2+6+7)
2x2=42+2=4I tried negative numbers and didn't find any.
I cannot take this question seriously! The sum is 75.
3+2i + 6-4i = 9-2i The real part of this number is positive, therefore it lies in Q1 or Q4. The imaginary part is negative, therefore it is in Q3 or Q4. Q4 is the common possibility, therefore 9-2i is in Q4.
The sum of -8 and 6 is -2