It would be 8 minus 9i
or 8-9i
Just add the real part and the imaginary part separately.
To form the additive inverse, negate all parts of the complex number → 8 + 3i → -8 - 3i The sum of a number and its additive inverse is 0: (8 + 3i) + (-8 - 3i) = (8 + -8) + (3 + -3)i = (8 - 8) + (3 - 3)i = 0 + 0i = 0.
x=5
15b + 13c - 12b + 10c + 8 = 3b + 23c + 8
8 of anything plus 8 more of the same thing totals up to 16 of them.
8 - 8i
8
(7 + 3i) + (8 + 9i) = (7 + 8) + (3i + 9i) = (7 + 8) + (3 + 9)i = 15 + 12i Which can also be written as: 15 + 12i = 3(5 + 4i).
-6i-8
(8+6i)-(2+3i)=6+3i 8+6i-2+3i=6+9i
Since you didn't show an operator, we'll use: 1. 8-6i 2. 8+6i 3. 8 times 6i = 48i The complex conjugates are: 1. 8+6i 2. 8-6i 3. -48i
No.The roots are the complex conjugate pair 5 ± 2.4495iwhere i is the imaginary square root of -1.
12
113
The complex conjugate pair, 4 ± 3.7417i where iis imaginary square root of -1.
162
241