Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Possible Duplicate:
Complex number: calculate $(1 + i)^n$.

I came across a difficult problem which I would like to ask you about:

Compute $ (1+i)^n $ for $ n \in \mathbb{Z}$

My ideas so far were to write out what this expression gives for $n=1,2,\ldots,8$, but I see no pattern such that I can come up with a closed formula.

Then I remember that one can write any complex number $a+bi$ like:

$$(a+bi)=\sqrt{a^2+b^2} \cdot \left( \frac{a}{\sqrt{a^2+b^2}}+\frac{b}{\sqrt{a^2+b^2}}\cdot i\right)$$

and $\frac{a}{\sqrt{a^2+b^2}} = \cos(\phi)$ and $\frac{b}{\sqrt{a^2+b^2}} = \sin(\phi)$ where $\phi$ is $\arctan{\frac{b}{a}} $

So it becomes,

$(a+bi)=\sqrt{a^2+b^2} \cdot ( \cos(\phi)+\sin(\phi)\cdot i)$ Taking this entire thing to the power $n$ using De Moivre

$$(a+bi)^n=(\sqrt{a^2+b^2})^n \cdot ( \cos(n\phi)+\sin(n\phi)\cdot i)$$ Substituting my $a=1$ and $b=1$

$(1+i)^n=(\sqrt{2})^n \cdot ( \cos(n\cdot\frac{\pi}{4})+\sin(n\cdot\frac{\pi}{4})\cdot i)$

$\phi$ is 45 degrees hence $\frac{\pi}{4}$

But now I don't know how to continue further and I would really appreciate any help! Again, Im looking for a closed formula depending on n.

Best regards

share|improve this question
I think it would be a lot easier to observe that $(i+1)^2 = 2i$, so therefore $(i+1)^{2n} = (2i)^n$, and $(i+1)^{2n+1} = (2i)^n(i+1)$. No trigonometry or De Moivre's theorem is required here. –  MJD Sep 19 '12 at 7:18
Upvoted for a clear, well-researched question. Voted to close as duplicate. –  The Chaz 2.0 Sep 19 '12 at 13:36
add comment

marked as duplicate by lhf, The Chaz 2.0, MJD, Chris Eagle, J. M. Sep 20 '12 at 17:04

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

4 Answers

up vote 2 down vote accepted

I think you are making this much too difficult, because:

$$(1+i)^2 = 2i.$$

So $(1+i)^{2m} = (2i)^m = 2^mi^m$.

That takes care of the even powers of $1+i$.

For the odd powers, just multiply by $1+i$ again: $(1+i)^{2m+1} = (1+i)^{2m}(1+i) = 2^mi^m(i+1)$.

If you want to simplify further, you need to know the remainder when $m = \bigl\lfloor \frac n2 \bigr\rfloor$ is divided by 4, so that you can reduce the $i^m$ part.

Say $n$ is even, and $m=n/2 = 4k+j$, where $0\le j<3$; then $n=8k+2j$:

$$\begin{array}{ll} j & (1+i)^{n=8k+2j} \\ \hline 0 & \hphantom{-}2^m \\ 1 & \hphantom{-}2^mi \\ 2 & -2^m \\ 3 & -2^mi \end{array}$$

If $n$ is odd, and $m=(n-1)/2 = 4k+j$, where $0\le j<3$, then $n=8k+2j+1$:

$$\begin{array}{ll} j & (1+i)^{n=8k+2j+1} \\ \hline 0 & \hphantom{-}2^m+2^mi \\ 1 & -2^m+2^mi \\ 2 & -2^m-2^mi \\ 3 & \hphantom{-}2^m-2^mi \end{array}$$

Putting these together into one table, we can just say that $n=8k+j'$, where $0\le j'\lt 8$:

$$\begin{array}{lrl} j' & (1+i)^{n=8k+j'} \\ \hline 0 & \hphantom{-}2^m & \\ 1 & \hphantom{-}2^m&+2^mi \\ 2 & &\hphantom{-}2^mi \\ 3 & -2^m&+2^mi \\ 4 & -2^m& \\ 5 & -2^m&-2^mi \\ 6 & &-2^mi \\ 7 & \hphantom{-}2^m&-2^mi \end{array}$$

(And remember $m=\bigl\lfloor \frac n2 \bigr\rfloor$.)

share|improve this answer
add comment

You got to this expression: $$(1+i)^n=\left(\sqrt{2}\right)^n \left(\cos\left(n\frac{\pi}{4}\right)+i\sin\left(n\frac{\pi}{4}\right)\right)$$ Now, you should note that this expression (without $\left(\sqrt{2}\right)^n$) has periodicity of $8$, i.e. for $n=k$ and $n=k+8$ the value will be the same. So you can just look at $n=0,1,...,7$ and write the answer in the form: $$(1+i)^n=\left(\sqrt{2}\right)^n\left\{ \begin{array}{cc} 1 & n\equiv0(\mod8)\\ \vdots & \end{array}\right.$$

share|improve this answer
Like your way here (+1). –  Basil R Sep 19 '12 at 9:19
add comment

You have a closed form already if you do just a little more work:

$$\begin{align*} (1+i)^n&=\left(\sqrt{2}\right)^n\left(\cos\frac{n\pi}4+i\sin\frac{n\pi}4\right)\\ &=2^{n/2}\left(\cos\frac{n\pi}4+i\sin\frac{n\pi}4\right)\\ &=\begin{cases} 2^{n/2},&\text{if }n\equiv 0\pmod 8\\ 2^{n/2}\left(\frac{\sqrt2}2+\frac{\sqrt2}2i\right),&\text{if }n\equiv 1\pmod 8\\ 2^{n/2}i,&\text{if }n\equiv 2\pmod 8\\ 2^{n/2}\left(\frac{\sqrt2}2-\frac{\sqrt2}2i\right),&\text{if }n\equiv 3\pmod 8\\ -2^{n/2},&\text{if }n\equiv 4\pmod 8\\ -2^{n/2}\left(\frac{\sqrt2}2+\frac{\sqrt2}2i\right),&\text{if }n\equiv 5\pmod 8\\ -2^{n/2}i,&\text{if }n\equiv 6\pmod 8\\ -2^{n/2}\left(\frac{\sqrt2}2-\frac{\sqrt2}2i\right),&\text{if }n\equiv 7\pmod 8\;. \end{cases}\\ \end{align*}$$

A function defined by cases is still a closed form. You can do better, though, if you allow the exponential form of the complex number: $1+i=\sqrt2 e^{i\pi/4}$, so


share|improve this answer
add comment

Use: $$\cos(n\theta)=\cos^n(\theta)-\frac{n(n-1)}{2!}\cos^{n-2}(\theta)\sin^2(\theta)+\frac{n(n-1)(n-2)(n-3)}{4!}\cos^{n-4}(\theta)\sin^4(\theta)-...$$ and $$\sin(n\theta)=n\cos^{n-1}(\theta)\sin(\theta)-\frac{n(n-1)(n-2)}{3!}\cos^{n-3}(\theta)\sin^3(\theta)-...$$

share|improve this answer
Nice! well-done, my friend, and hoping you're reading this as your new day begins! –  amWhy Mar 21 '13 at 1:48
@amWhy: Thanks; So GN to me. ;-) –  B. S. Mar 21 '13 at 20:25
add comment

Not the answer you're looking for? Browse other questions tagged or ask your own question.