Evaluate $\dfrac{(1+i)^{n+2}}{(1-i)^n}$
I think that the meaning is that it need to be simplified. Thanks
Asked
Active
Viewed 97 times
4
-
5Hint: $\dfrac{1+i}{1-i} = \dfrac{2i}{(1-i)(1+i)} = i$, $(1+i)^2 = 2i$. – njguliyev Oct 26 '13 at 21:47
-
So because the two expressions are in the power of n, the result will include (1-i)power 2n? Thanks! – gbox Oct 27 '13 at 18:32
-
I don't understand your question, but $$\dfrac{(1+i)^{n+2}}{(1-i)^n} = \left(\dfrac{1+i}{1-i}\right)^n\cdot(1+i)^2.$$ – njguliyev Oct 27 '13 at 21:21
-
What I mean is that $\dfrac{(1+i)^{n+2}}{(1-i)^n}$ needs to be multiple by is z* or $\dfrac{(1+i)^{n}}{(1+i)^n}$ That gives $\dfrac{(1+i)^{2n+2}}$ – gbox Oct 28 '13 at 05:34
2 Answers
1
$$\dfrac{(1+i)^{n+2}}{(1-i)^n}=\frac{2i(1+i)^n}{(1-i)^n}=2i\frac{(1+i)^n}{(1-i)^n}=2i(-i)^{-n}=2ie^{\frac{1}{2}\pi in}$$
Jan Eerland
- 28,671
-1
This simplifies to $2 i^{n+1}$.
Riemann1337
- 604
- 4
- 10
-
5The OP is probably more interested in the process of the simplification- not just the final answer. – Sujaan Kunalan Oct 26 '13 at 21:48