How is Complex z Raised to a Power Equal to Exponential?

[naggy]

How can this be true(z is complex)

z^a = exp(a * ln(z))

but in general ln(z^a) is not equal to a * ln(z)

 

[jostpuur]

Are you mainly interested in the case where a\in\mathbb{R}? (Answer would be simpler then)

 

[jostpuur]

Well, anyway, I'll assume that a\in\mathbb{R}. Arbitrary a\in\mathbb{C} would result in more calculating, but I don't believe there would be anything highly different.

How can this be true(z is complex)

z^a = exp(a * ln(z))

This equation is usually true, because it is the definition of the left side. First you must decide what argument function \textrm{arg}:\mathbb{C}\to\mathbb{R} you are using. This choice will fix the branch of the logarithm,

<br /> \log(z) = \log |z| + i\textrm{arg}(z),<br />

and then we denote

<br /> z^a := e^{a\log (z)}.<br />

The equation is not necessarily true, if you start using other branches of logarithm later, of course.

but in general ln(z^a) is not equal to a * ln(z)

This is not true always because of the branch cuts.

<br /> \log (z^a) = \log(e^{a\log(z)}) = \log( \underbrace{e^{a\log |z|}}_{\in\mathbb{R}} \underbrace{e^{ia\textrm{arg}(z)}}_{\in S^1\subset\mathbb{C}} ) = \log(e^{a\log |z|}) \;+\; i\textrm{arg}(e^{ia\textrm{arg}(z)})<br />

<br /> a\log(z) = a\log |z| + ia\textrm{arg}(z)<br />

The equation

<br /> a\log |z| = \log(e^{a\log |z|})<br />

is always true, but

<br /> a\textrm{arg}(z) = \textrm{arg}(e^{ia\textrm{arg}(z)}) + 2\pi n<br />

may need some n\in\mathbb{Z} depending on how the argument function happens to behave with these a,z.

 

[naggy]

Well, anyway, I'll assume that a\in\mathbb{R}. Arbitrary a\in\mathbb{C} would result in more calculating, but I don't believe there would be anything highly different.



This equation is usually true, because it is the definition of the left side. First you must decide what argument function \textrm{arg}:\mathbb{C}\to\mathbb{R} you are using. This choice will fix the branch of the logarithm,

<br /> \log(z) = \log |z| + i\textrm{arg}(z),<br />

and then we denote

<br /> z^a := e^{a\log (z)}.<br />

The equation is not necessarily true, if you start using other branches of logarithm later, of course.



This is not true always because of the branch cuts.

<br /> \log (z^a) = \log(e^{a\log(z)}) = \log( \underbrace{e^{a\log |z|}}_{\in\mathbb{R}} \underbrace{e^{ia\textrm{arg}(z)}}_{\in S^1\subset\mathbb{C}} ) = \log(e^{a\log |z|}) \;+\; i\textrm{arg}(e^{ia\textrm{arg}(z)})<br />

<br /> a\log(z) = a\log |z| + ia\textrm{arg}(z)<br />

The equation

<br /> a\log |z| = \log(e^{a\log |z|})<br />

is always true, but

<br /> a\textrm{arg}(z) = \textrm{arg}(e^{ia\textrm{arg}(z)}) + 2\pi n<br />

may need some n\in\mathbb{Z} depending on how the argument function happens to behave with these a,z.

So... basically it's not true due to the extra 2*pi*n factor at the end?

No that can´t be true. Then the equation would be true for the principal value?

No wait. It's the extra iarg factor in this

<br /> \log (z^a) = \log(e^{a\log(z)}) = \log( \underbrace{e^{a\log |z|}}_{\in\mathbb{R}} \underbrace{e^{ia\textrm{arg}(z)}}_{\in S^1\subset\mathbb{C}} ) = \log(e^{a\log |z|}) \;+\; i\textrm{arg}(e^{ia\textrm{arg}(z)})<br />

If there is no iarg, then there is only log(e^(a*log|z|)) which is |z|^a and this is true for the real numbers since |z| is really replaced by x