# [SOLVED] inductive proof

#### epreble

for n in the natural numbers prove that (2+i)^n is never real.

base case
(2+i) is not real

inductive hypothesis
for k=n assume (2+i)^k is not in the reals
must prove
(2+i)^(k+1)

Ive got to the point where
2(2+i)^k + i(2+i)^k must not be real. its seems logical that the i terms would not cancel out making this sum not real but i cant figure out a way to prove it.
I've tried using the binomial theorem. Any HINTS would be greatly appreciated.

#### Defunkt

MHF Hall of Honor
If $$\displaystyle (2+i)^k$$ is not real then you can write it as $$\displaystyle a + i \cdot b$$ for $$\displaystyle a, b \in \mathbb{R}$$. Now, what is $$\displaystyle (2+i)(2+i)^k$$ ?

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#### epreble

(2+i)((2+i)^k) = (2+i)^(k+1)

#### Defunkt

MHF Hall of Honor
Correct. Now use my hint to prove the inductive step.

epreble

#### epreble

$$\displaystyle (2+i)(2+i)^k$$ = (a+ib)(2+i)