$1^x = i$
I can't solve it through logs, because $\log 1 = 0$. Does this mean $x$ is undefined?
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4$i^4$ = 1, but remember that in complex domain there are 4 roots of degree four, so your problem is not well-posed. – dtldarek May 19 '12 at 12:44
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What would it mean when $1^x=-1$? Impossible Farhad? – Mikasa May 19 '12 at 13:00
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As already mentioned in the comments, your problem is not well-posed. Since $i^4 = 1$, you would like to say that $1^{1/4} = i$, but this is not entirely true. In the complex domain, the root function, such as $\sqrt[4]{\cdot}$ is not a well-defined function, because it in general has 4 different values for a single input. It is what is called a multivalued function, or some variant of this.
Rick
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$(\log 1) * x = \log i$. $0 * x = \log i$. $x = \frac{\log i}{0}$Anything divided by zero is always undefined, but as you proved one possible value is 1/4. Isn't this the equivalent of giving a value to an undefined equation? Or is this simply saying something divided by zero is a multivalued function rather than undefined? – May 19 '12 at 13:23
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- The statement $(\log 1) x = \log i$ is incorrect, again for the same reason: the logarithm function $\log$ is not a well-defined function, because for a single input it outputs (infinitely) many different values. 2) The statement "Anything divided by zero is always undefined" is incorrect; it is correct to say that "dividing by zero is an undefined operation," i.e. "dividing by zero is forbidden -- such an operation does simply not exist!"
– Rick May 19 '12 at 13:46 -
In short, even (erroneously) assuming that the equation $1^x = i$ is well-posed, it cannot be solved using logarithms. – Rick May 19 '12 at 13:47
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3Although, of course, assuming something false can imply anything you wish... so, on a philosophical level, assuming $1^x = i$ is well-posed, it can be solved using logarithms. As well as bunnies. – Rick May 19 '12 at 13:49
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It's true that logarithms are multivalued functions. What this means is that you won't get a unique answer. However certain numbers will clearly work while others do not. What you can do to find solutions is this:
$$1^x=i$$ $$x\ln(1)=\ln(i)$$
We need to find complex values of both of these logarithms. In the following, $n \in \mathbb{Z}$.
$$\ln(1)=a+bi$$ $$1=e^{a+bi}$$ $$1=e^a(\cos(b)+i\sin(b))$$
The imaginary component must be 0 and the real component must be 1. This implies that $$b=\pi n$$ for any n. This becomes $$1=e^a(\cos(\pi n)+i\sin(\pi n))$$ $$1=e^a(-1)^n$$
From here it's clear that $a=0$ and $n$ is even, so the full solution for $\ln(1)$ is $$ln(1)=2\pi n i$$ Similarly, with $\ln(i)$: $$i=e^a(\cos(b)+i\sin(b))$$ Now the real part must be zero $$b=\frac{\pi (2n+1)}{2}$$ $$i=ie^a(-1)^n$$ So again, $a=0$ and $n$ is even. $$\ln(i)=\frac{\pi(4n+1)}{2}i$$
Subbing into your equation:
$$x\ln(1)=ln(i)$$ $$x\cdot 2\pi n i=\frac{\pi(4m+1)}{2}i$$ with $m \in \mathbb{Z}$ as well. Solving gives: $$x=\frac{4m+1}{4n}$$ for any $m,n \in \mathbb{Z}, n \neq 0$.
Saying $\ln(x)$ is undefined in the complex plane is like saying $\sqrt x$ is undefined on the reals. 'Undefined' doesn't mean that no solutions exist, but that multiple solutions exist, meaning that it isn't a function anymore ('multivalued function' is a misnomer since functions by definition only have one output value for a given input). We make it a function by defining a 'principal branch'. For $\sqrt x$ we do this by only considering the positive branch. For complex $\ln(x)$ we do this by restricting the imaginary part of the answer to be in$(-\pi,\pi]$. This would involve dividing by zero here, meaning that the only solutions to your question involve considering non-principal values of the logarithm.
Robert Mastragostino
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