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I have to solve the following complex number exercise: calculate $(1 + i)^n\forall n\in\mathbb{N}$ giving the result in $a + ib$ notation.

Basically what I have done is calculate $(1 + i)^n$ for some $n$ values. $$(1 + i)^1 = 1 + i$$ $$(1 + i)^2 = 2i$$ $$(1 + i)^3 = - 2 + 2i$$ $$\boxed{(1 + i)^4 = - 4}$$ $$(1 + i)^5 = (1 + i)^4\cdot(1 + i)^1 = (-4)\cdot(1 + i) = - 4 - 4i$$ $$(1 + i)^6 = (1 + i)^4\cdot(1 + i)^2 = (-4)\cdot2i = - 8i$$ $$(1 + i)^7 = (1 + i)^4\cdot(1 + i)^3 = (-4)\cdot(- 2 + 2i) = 8 - 8i$$ $$(1 + i)^8 = (1 + i)^4\cdot(1 + i)^4 = (-4)\cdot(-4) = (-4)^2 = 16$$ We can write $n = 4\cdot q + r$ (Euclidean division), so we have: $$(1 + i)^n = (1 + i)^{(4\cdot q + r)} = ((1 + i)^4)^q\cdot(1 + i)^r = (-4)^q\cdot(1 + i)^r$$ Finally if you want to calculate say... $(1 + i)^n$ for $n = 625$ you have: $$625 = 4\cdot156 + 1\Rightarrow q = 156, r = 1$$ $$(1 + i)^{625} = (-4)^{156}\cdot(1 + i)^1 = (-4)^{156} + (-4)^{156}i$$ What other approach would you suggest? Mine works, but you have to find $q$ and $r$ in order to do the calculation, and I think it is not "calculate" technically speaking which was what the exercise is asking (although I am not sure what they mean by calculate).

Adelaide Dokras
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David Robert Jones
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    Hint: USe Euler's identity to write $1+i$ in the form $re^{i\theta}$ for suitably chosen $r$ and $\theta$. Then, $(1+i)^n = (re^{i\theta})^n = r^ne^{in\theta}$. – Dilip Sarwate May 25 '12 at 20:28
  • This is related. – Did May 25 '12 at 20:30
  • It is a calculation, you can note the pattern, come up with a general formula, and then prove it (technically by induction) in much the same way as you dealt with $625$. But the polar form $re^{i\theta}=r(\cos\theta +i\sin\theta)$ is much more useful. – André Nicolas May 25 '12 at 20:32
  • If, as your comments suggest, de Moivre's formula is "cheating", find a recurrence relation for the successive coefficients of 1 and i in $(1+i)^n$ i.e. real and imaginary parts, and solve that. The thing is completely equivalent. If this is part of an exercise to motivate de Moivre and prove it, that is one thing. But if you know de Moivre, why not use it? – Mark Bennet May 25 '12 at 20:49
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    For my money, in agreement with @Hurkyl, I think the solution you’ve given is best possible, being direct and transparent. The fact that the form of the answer depends on the congruence class of $n$ modulo $8$ is not a disadvantage, but a fundamental aspect of the problem. You evidently wanted a closed-form answer, but what you got is plenty good enough. – Lubin May 26 '12 at 00:50

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Write $1+i$ in modulus argument notation ($\rho e^{i \theta}$). Then $(1+i)^n = \rho^n e^{i n \theta}$ will be pretty easy to compute.

Albert
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  • You are right, but they want it in a + ib notation. Modulus argument notation is straightforward, but it's part B of the exercise :) – David Robert Jones May 25 '12 at 20:31
  • As Glouglo. noted write $1+i$ as $\rho e^{i\theta}$ and then you can use de Moivre's formula (http://en.wikipedia.org/wiki/De_Moivre's_formula). – Mikasa May 25 '12 at 20:35
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    OK, and then write nack $\rho^n e^{i n \theta} = \rho^n (\cos(n \theta) + i \sin(n \theta))$. – Albert May 25 '12 at 20:43
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Perhaps, proceed via the trigonometric form using De Moivre's formula: $$\left(1+i\right)^{n}=2^{\frac{n}{2}}\left(\cos\frac{\pi}{4}+i\sin\frac{\pi}{4}\right)^{n}=2^{\frac{n}{2}}\cos\frac{\pi n}{4}+i2^{\frac{n}{2}}\sin\frac{\pi n}{4}$$

Valentin
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Fair enough, you could then try binomial expansion $$\left(1+i\right)^{n}=\sum_{k=0}^{n}\left(\begin{array}{c}n\\k \end{array}\right)1^{k}i^{n-k}=\sum_{k=0}^{n}\left(\begin{array}{c} n\\k\end{array}\right)i^{n-k}=\sum_{k=0}^{n}\left(\begin{array}{c}n\\n-k \end{array}\right)i^{k}$$ (changing the summation variable $n-k\to k$ in the last transition. Then use the cyclic property of $i$ which you implicitly relied upon in your solution: $$i^{2}=-1$$ $$i^{3}=-i$$ $$i^{4}=1$$ $$i^{5}=i$$ to break down the sum in real and imaginary parts. If the exercise does not require a "closed" form, you probably would not even need to bother about combinatorial identities.

Valentin
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Mine works, but you have to find q and r in order to do the calculation,
and I think it is not "calculate" technically speaking which was what the
exercise is asking (although I am not sure what they mean by calculate).

To be honest, the form you worked out is probably the best form of the answer for the specific question. Although other forms involving $e^{i \theta}$ or the trigonometric functions are snazzier and more succinct, they will typically be less practical to use except in special circumstances.

Using the polar form of the complex number (or de Moivre's formula) is probably the simplest way to derive your answer, of course. But the way you did it is an eminently reasonable method; it's simple and straightforward, and it's only real drawbacks are that it's more 'messy' and doesn't generalize well.

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How about setting L = lim_n/rarrow (1 + i) ^n

DaveUM
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  • Sorry, I still haven't mastered the formatting you'all use so again I must resort to prose. Can someone take me by the hand with your format editor? – DaveUM Jan 26 '13 at 15:03
  • To the point of the problem. If you set L = the original limit and take the log of both sides. The n will come down to "ground zero" and easier to deal with. After computing the new limit remember to resolve by reversing the log(L) = A to L=e^A. – DaveUM Jan 26 '13 at 15:07
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    I don't see how this answers the problem. Rather than computing a power of a complex number, it is taking a limit of such powers. – rschwieb Jan 26 '13 at 15:51