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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).

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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.2012-05-26

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

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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.2013-01-26
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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.

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    OK, and then write nack $\rho^n e^{i n \theta} = \rho^n (\cos(n \theta) + i \sin(n \theta))$.2012-05-25
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Perhaps, proceed via the trigonometric form using De Moivre's formula: $\left(1+i\right)^{n}=2^{\frac{n}{2}}\left(\cos\frac{\pi}{4}+i\sin\frac{\pi}{4}\right)^{n}=2^{\frac{n}{2}}\cos\frac{\pi n}{4}+i2^{\frac{n}{2}}\sin\frac{\pi n}{4}$

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    Using De Moivre's formula is like cheating? Why!?2012-05-25
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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.

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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.