3 to the power of 6 is 729.
- 2 - 3i
s3 - 3s2 + 9s - 27 = s2(s - 3) + 9(s - 3) = (s - 3)(s2 + 9) or = (s - 3)(s2 - (9i2) = (s - 3)(s - 3i)(s + 3i) if you want to find for what values of s the expression equals to zero, then this happens when s - 3 = 0 or s - 3i = 0 or s + 3i = 0 s = 3, ±3i
0 + 3i
11
The complex conjugate of 2-3i is 2+3i.
I would certainly bet on a computer. Try to solve (log z)^-3i. This is log z raised to the -3i power. A computer comes in handy.
(8+6i)-(2+3i)=6+3i 8+6i-2+3i=6+9i
-2 - 3i
(3i+l)(3-I)
The "i" in 3i means the number is imaginary.
0+3i has a complex conjugate of 0-3i thus when you multiply them together (0+3i)(0-3i)= 0-9i2 i2= -1 0--9 = 0+9 =9 conjugates are used to eliminate the imaginary parts