is $1^z=1$ for all complex values of $z$?

Question

i would like to see if $1^z=1$ is valid for all complex variable $z$,first of all you can rewrite above equation as

$1^{a+b*i}=e^0$ here i think that instead of $+$ sign, we may take take complex conjugate form or $-$ sign.from above equation we can get

$1^a *1^{b*i}=e^0$

or $1^{b*i}=1$

here i am assuming that $a$ is real,otherwise if we have complex variable in power complex variable,result will be undefined,so now question is: could you conclude that

$1^{b*i}=1$?

generally i know that $i^0=1$

so

$i^{0*b*i}=1$

we get identity ,but is this right?

i would like to show you following article http://www.cut-the-knot.org/do_you_know/complex.shtml

with title : Complex number to a complex power may be real

what could i say above equation? also we may introduce some useful well known identities,but which one could be relevant for this case?thanks very much

score 2 · Accepted Answer · answered Apr 09 '13 at 19:54

2

It depends on the particular branch of the complex logarithm you choose, for example:

$$1^i=e^{i\operatorname{Log}1}=e^{i\log|1|-\arg 1}=e^{-\arg 1}\in\{e^0=1\;,\;e^{-2\pi}\;,\;e^{2\pi}\;,\;e^{-2\pi}\,,\ldots\}$$

answered Apr 09 '13 at 19:54

DonAntonio

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so one necessary condition for equation hold is that in $z=a+b*i$ , $b=0$? or complex number contains only real part? – dato datuashvili Apr 09 '13 at 19:58
Well, in this case it is necessary and sufficient, as you can see. I can't understand your question "or complex number contains only real part?" – DonAntonio Apr 09 '13 at 19:59
sorry i should ask it to Three,wrong place,sorry.i have understood it,just when he said unless $b=0$,i have asked about it.when imaginary part is zero,clearly complex number contains only real part i meant this.thanks – dato datuashvili Apr 09 '13 at 20:00
i have one question @DonAntonio,what is arg1 stand for?is it just some number or argument of complex number? – dato datuashvili Apr 09 '13 at 20:06
argument, @dato – DonAntonio Apr 09 '13 at 21:37

score 1 · Answer 2 · answered Apr 09 '13 at 19:54

1

$ 1^{ib}=({e^{2\pi i }})^{ib}= e^{-2\pi b}\neq1 $, unless $ b = 0$

Here, I've used the fact that $e^{2\pi i} = 1 $ and $ i^2 = -1$

answered Apr 09 '13 at 19:54

Three

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Here, for example, Three chose the branch of the complex logarithm for which $,\arg(1)=2\pi,$ ... – DonAntonio Apr 09 '13 at 19:55

is $1^z=1$ for all complex values of $z$?

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