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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$\begingroup$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.
$\endgroup$ 3 $\begingroup$Perhaps, proceed via the trigonometric form using De Moivre's formula: $$\left(1+i\right)^{n}=2^{\frac{n}{2}}\left(\cos\frac{\pi}{4}+i\sin\frac{\pi}{4}\right)^{n}=2^{\frac{n}{2}}\cos\frac{\pi n}{4}+i2^{\frac{n}{2}}\sin\frac{\pi n}{4}$$
$\endgroup$ 2 $\begingroup$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.
$\endgroup$ $\begingroup$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.
$\endgroup$ $\begingroup$How about setting L = lim_n/rarrow (1 + i) ^n
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