-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.
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.
The sum of -8 and 6 is -2
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.
Do this in reverse. The sum of -1 and -1 is -1+-1=-2 The difference of -6 and -6 is -6-(-6)=-6+6=0 0 increased by -2 is 0+-2=-2 The sum of 10 and -2 is 10+-2=8 Translation: 8