nth distinct root

Question

How do I show (1+i)^n = -9 algebraically and graphically.

I figured out that polar form form 1+i. Not sure what to do with -9

sqrt 2^p(cos (p⋅45°)+ i sin (p⋅45°).

please show work

Answer 1

1+I to the 2 power = (1+I)^2 = (1+I)(1+I) = 1 +2i -1 = 2i

1+I to the 4 power = (2i)^2 = -4

1+I to the 8 power = (-4)^2 = 16

1+I to the 9 power = 16(1+I) = 16+16i

1+I to the 6 power = (-4)(2i) = -8i

1+I to the 5 power = -4(1+I) = -4 -4i

1+I to the 3 power = 2i(1+I) = 2i-2

1+I to the 7 power = -8i(1+I) = -8k +8

no value of n will make (1+I)^n = -9

but you could take natural logs

nln(1+I) = ln(-9)

n = (ln(-9))/ln(1+I) but you can't take a log of a negative number like -9

graph (1+I)^x= -9 and nothing shows up on a graphing calculator

Answer 2

Well I calculated the powers of 1 + i from 2 through 8 and none of them result in -9.

Please check your post for accuracy.