What is the value of $i^0$?
I have to solve the following question -
$$\sum_{n=0}^{1000} i^n$$
where $i = \sqrt{-1}$
To be able to solve the problem, I need to know the value of $i^0$. What is the value of $i^0$? Is it 0 or indeterminate or something else?
$\endgroup$ 117 Answers
$\begingroup$Just use the same approach as with (one-dimensional) real numbers:
$i^3=-i$
$i^2=-1$
$i^1=i$
To go up the list above, multiply by $i$. To go down the list, we do the opposite and divide by $i$. $i^0$, then, must logically be the next element below, which means
$i^0=\frac{i^1}{i}=1$
$\endgroup$ 2 $\begingroup$In every multiplicative group $G$ we define $$ g^0=e $$ for every $g\in G$. In this case we have $G=\mathbb C^*$
One can use the definition $z^a=e^{a\log(z)}$, in this case we defined $\log(z)$ for all complex numbers except negative real numbers. And $e^z$ is defined for every $z\in\mathbb C$, so by this definition we have $$ i^0=1 $$
$\endgroup$ 11 $\begingroup$We have $x^0:= 1$ for every complex number $x$. (Notice that this is the only convention which fits into the rules of arithmetic, and there is no need to exclude $x=0$. Think about the binomial theorem, for instance.)
By the way, your exercise $\sum_{n=0}^{1000} i^n$ can be solved with the usual formula for geometric series. Thus, it is $\dfrac{i^{1001}-1}{i-1}$ and now you can simplify this using $i^k = i^{k \bmod 4}$.
$\endgroup$ 20 $\begingroup$For any complex number $z\not = 0$, we have $z^0 = 1$. There are two reasons:
1) By definition, a nonzero complex number $z = re^{i\theta}$ has $z^n = r^n e^{i n\theta}$ for any integer $n$. Note that since $z\not = 0$, the value of $\theta$ is determined modulo $2\pi$. (In fact, the reason that this definition only works for $n$ an integer is that $\theta$ is only defined modulo $2\pi$. Then $2\pi n$ is also a multiple of $2\pi$, so $r^n e^{in\theta}$ is still well-defined: $e^{in\theta} = e^{in(\theta + 2\pi)}$.) This definition works perfectly well for $n = 0$.
2) We want $z^{n+m} = z^n z^m$ for integers $n, m$ and $z\not = 0$. In particular, we want $z = z^{1+0} = z^0 z^1 = (z^0) z$, which holds iff $z^0 = 1$. Again, this argument works for any nonzero complex $z$, including $z = i$.
$\endgroup$ $\begingroup$Provided that you know Euler's representation of complex numbers: $$i:=e^{i\pi/2}$$ So...
EDIT: There is no recursive definition here. Euler's representation $re^{i\theta}$ refers to the point $(r,\theta)$ (expressed in the polar coordinate system) on the complex plane, and hence you can well forget the square root definition when you work within Euler's system.
EDIT 2: And if you are seeking the reason why the two ways of defining $i$ are equivalent, you could grasp a shallow perspective by taking a glance at the Dimov's formula. But, as to the deep reason behind that, I have to confess, it's beyond what I could tell.
EDIT 3: Ultimately, you have to understand that $i^0=1$ is also a definition (as is pointed out in that group theory answer). We define it this way so as to make the notation of the computation of complex numbers as compatible with what we do within real numbers as possible. $\endgroup$ 3 $\begingroup$
From the rules of indices $i^{0}=1$
$\endgroup$ 2 $\begingroup$$$i^0=i^{1-1}=\frac{i^1}{i^1}=\frac ii$$ Now, $\frac ii$ is asking us for the number $c$ such that $c\times i=i$. It's not hard to show that $c=1$ is the only one that works. Thus, $\frac ii=1$, and: $$i^0=1$$ (This exact argument shows that $z^0=1$ for any nonzero number $z$.)
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