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3=(1+i)^15

3^(1/15)-1=i

(1+i)^ 15 = 3 is FALSE:

Simplify the following:
(1+i)^15

(1+i)^15 = (1+i) (1+i)^14 = (1+i) ((1+i)^7)^2 = (1+i) ((1+i) (1+i)^6)^2 = (1+i) ((1+i) ((1+i)^3)^2)^2 = (1+i) ((1+i) ((1+i) (1+i)^2)^2)^2:
(1+i) ((1+i) ((1+i) (1+i)^2)^2)^2

(1+i)^2 = 1+0+1 i+0+1 i-1 = 2 i:
(1+i) ((1+i) ((1+i)×2 i)^2)^2

i (1+i) = -1+i:
(1+i) ((1+i) (i-1 2)^2)^2

((i-1)×2)^2 = 2^2 (i-1)^2:
(1+i) ((1+i)×2^2 (i-1)^2)^2

2^2 = 4:
(1+i) ((1+i)×4 (i-1)^2)^2

(i-1)^2 = 1+0-i+0-i-1 = -2 i:
(1+i) ((1+i)×4×-2 i)^2

4 (-2) = -8:
(1+i) ((1+i)×-8 i)^2

i (1+i) = -1+i:
(1+i) (-8i-1)^2

((i-1) (-8))^2 = (-8)^2 (i-1)^2:
(1+i)×(-8)^2 (i-1)^2

(-8)^2 = 64:
(1+i)×64 (i-1)^2

(i-1)^2 = 1+0-i+0-i-1 = -2 i:
(1+i)×64×-2 i

64 (-2) = -128:
(1+i)×-128 i

i (1+i) = -1+i:
Answer: | 128 - 128i
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Hi guest

i is not an imaginary number it stands for interest.

This equaton is for determining the rate of compond interest.