How would you solve that problem? Would you add the exponents together then solve? Or determine the value of the i first?
2007-10-03 08:18:40 · 4 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
(4i)^3 = (4)^3 * (i)^3 = 64 * -i = -64i
(2i)^2 = (2)^2 * (i)^2 = 4 * -1 = -4
-64i * -4 = 256i
i = sqrt(-1)
2007-10-03 08:25:24 · answer #1 · answered by PMP 5 · 0⤊ 0⤋
You need to specify the exact values to which the exponent is being applied. Did you mean (4*i)^3, or 4*i^3?
I'm going to assume you meant the former. Here's how you solve it:
(4*i)^3 * (2*i)^2
= 4^3*i^3 * 2^2*i^2
= (4^3*2^2)*(i^3*i^2)
= (64*4)*(i^5)
= 256*i^5
2007-10-03 15:27:10 · answer #2 · answered by lithiumdeuteride 7 · 0⤊ 0⤋
Unless things have changed since I have been in school isn't i = imaginary numbers. This would be where you are tyring to take the square root of a negative number. Is this what you are trying to do?
As far as what you asked. When you are multiplying numbers with exponents you add them, then solve.
2007-10-03 15:25:01 · answer #3 · answered by TGAL1701 2 · 0⤊ 0⤋
(4i)^3 * (2i)^2
64(i^3) * 4(i^2)
256i^5 is the answer
You multiply it out, and then add exponents.
That is if i is a nomral variable
2007-10-03 15:25:54 · answer #4 · answered by mark p 2 · 0⤊ 0⤋