On complex powers of complex numbers
$begingroup$
For $z,s in mathbb{C}$ and $zneq 0$, set $z^s = exp(s,log z)$ and $-pi < arg z leq pi$. In this setting, I am worried about the cases where I have to be careful in assuming $(z_1z_2)^s = (z_1)^s(z_2)^s$. As long as $s in mathbb{Z}$, I am safe, but what about the rest? For example, if $zin mathbb{H}$, then
begin{equation}
(-z)^s = e^{-pi is}z^s,
end{equation} but
begin{equation}
(-1)^sz^s = e^{pi i s}z^s
end{equation} and both are not equal.
complex-analysis complex-numbers modular-forms
edited Dec 17 '18 at 4:18
Tianlalu
3,08621038
asked Dec 17 '18 at 4:14
user166305user166305
1968
$endgroup$
add a comment |
$begingroup$
For $z,s in mathbb{C}$ and $zneq 0$, set $z^s = exp(s,log z)$ and $-pi < arg z leq pi$. In this setting, I am worried about the cases where I have to be careful in assuming $(z_1z_2)^s = (z_1)^s(z_2)^s$. As long as $s in mathbb{Z}$, I am safe, but what about the rest? For example, if $zin mathbb{H}$, then
begin{equation}
(-z)^s = e^{-pi is}z^s,
end{equation} but
begin{equation}
(-1)^sz^s = e^{pi i s}z^s
end{equation} and both are not equal.
complex-analysis complex-numbers modular-forms
edited Dec 17 '18 at 4:18
Tianlalu
3,08621038
asked Dec 17 '18 at 4:14
user166305user166305
1968
$endgroup$
add a comment |
$begingroup$
For $z,s in mathbb{C}$ and $zneq 0$, set $z^s = exp(s,log z)$ and $-pi < arg z leq pi$. In this setting, I am worried about the cases where I have to be careful in assuming $(z_1z_2)^s = (z_1)^s(z_2)^s$. As long as $s in mathbb{Z}$, I am safe, but what about the rest? For example, if $zin mathbb{H}$, then
begin{equation}
(-z)^s = e^{-pi is}z^s,
end{equation} but
begin{equation}
(-1)^sz^s = e^{pi i s}z^s
end{equation} and both are not equal.
complex-analysis complex-numbers modular-forms
edited Dec 17 '18 at 4:18
Tianlalu
3,08621038
asked Dec 17 '18 at 4:14
user166305user166305
1968
$endgroup$
For $z,s in mathbb{C}$ and $zneq 0$, set $z^s = exp(s,log z)$ and $-pi < arg z leq pi$. In this setting, I am worried about the cases where I have to be careful in assuming $(z_1z_2)^s = (z_1)^s(z_2)^s$. As long as $s in mathbb{Z}$, I am safe, but what about the rest? For example, if $zin mathbb{H}$, then
begin{equation}
(-z)^s = e^{-pi is}z^s,
end{equation} but
begin{equation}
(-1)^sz^s = e^{pi i s}z^s
end{equation} and both are not equal.
complex-analysis complex-numbers modular-forms
complex-analysis complex-numbers modular-forms
edited Dec 17 '18 at 4:18
Tianlalu
3,08621038
asked Dec 17 '18 at 4:14
user166305user166305
1968
edited Dec 17 '18 at 4:18
Tianlalu
3,08621038
asked Dec 17 '18 at 4:14
user166305user166305
1968
edited Dec 17 '18 at 4:18
Tianlalu
3,08621038
edited Dec 17 '18 at 4:18
Tianlalu
3,08621038
edited Dec 17 '18 at 4:18
Tianlalu
3,08621038
3,08621038
asked Dec 17 '18 at 4:14
user166305user166305
1968
asked Dec 17 '18 at 4:14
user166305user166305
1968
asked Dec 17 '18 at 4:14
user166305user166305
1968
1968
add a comment |
add a comment |
1 Answer
1
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$begingroup$
The rule $(zw)^s=z^sw^s$ depends on the equality $log(zw)=log z+log w$. But you need the logarithm to be single valued, which requires limiting the argument to one interval, for instance $(-pi,pi]$ as you did; and this generates the following problem. If $t+u>pi$, then $log (e^{i(t+u)})=i(t+u-2pi)$. Thus when $s=a+ib$, $z_j=r_je^{it_j}$,
begin{align}
(z_1z_2)^s&=exp((a+ib)(log r_1r_2+ i(t_1+t_2-2kpi))\ \
&=r_1^ar_2^a,e^{iblog r_1},e^{iblog r_2},e^{iat_1},e^{iat_2},e^{-bt_1},e^{-bt_2},e^{2bkpi},
end{align}
where $k=0$ if $t_1+t_2leqpi$, and $k=1$ when $t_1+t_2>pi$. On the other hand you have
$$
z_1^sz_2^s=r_1^ar_2^a,e^{iblog r_1},e^{iblog r_2},e^{iat_1},e^{iat_2},e^{-bt_1},e^{-bt_2}.
$$
The two expressions are equal when $t+uleqpi$, and differ by a factor of $e^{2bpi}$ when $t+u>pi$.
In summary, $(z_1z_2)^s=z_1^s+z_2^s$ (in your setup) when $arg z_1+arg z_2leqpi$. This is not terrible, but it prevents you from using the formula in general, when you don't know the arguments of $z_1$ and $z_2$.
answered Dec 17 '18 at 4:51
Martin ArgeramiMartin Argerami
125k1178178
$endgroup$
add a comment |
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1 Answer
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1 Answer
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active
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$begingroup$
The rule $(zw)^s=z^sw^s$ depends on the equality $log(zw)=log z+log w$. But you need the logarithm to be single valued, which requires limiting the argument to one interval, for instance $(-pi,pi]$ as you did; and this generates the following problem. If $t+u>pi$, then $log (e^{i(t+u)})=i(t+u-2pi)$. Thus when $s=a+ib$, $z_j=r_je^{it_j}$,
begin{align}
(z_1z_2)^s&=exp((a+ib)(log r_1r_2+ i(t_1+t_2-2kpi))\ \
&=r_1^ar_2^a,e^{iblog r_1},e^{iblog r_2},e^{iat_1},e^{iat_2},e^{-bt_1},e^{-bt_2},e^{2bkpi},
end{align}
where $k=0$ if $t_1+t_2leqpi$, and $k=1$ when $t_1+t_2>pi$. On the other hand you have
$$
z_1^sz_2^s=r_1^ar_2^a,e^{iblog r_1},e^{iblog r_2},e^{iat_1},e^{iat_2},e^{-bt_1},e^{-bt_2}.
$$
The two expressions are equal when $t+uleqpi$, and differ by a factor of $e^{2bpi}$ when $t+u>pi$.
In summary, $(z_1z_2)^s=z_1^s+z_2^s$ (in your setup) when $arg z_1+arg z_2leqpi$. This is not terrible, but it prevents you from using the formula in general, when you don't know the arguments of $z_1$ and $z_2$.
answered Dec 17 '18 at 4:51
Martin ArgeramiMartin Argerami
125k1178178
$endgroup$
add a comment |
$begingroup$
The rule $(zw)^s=z^sw^s$ depends on the equality $log(zw)=log z+log w$. But you need the logarithm to be single valued, which requires limiting the argument to one interval, for instance $(-pi,pi]$ as you did; and this generates the following problem. If $t+u>pi$, then $log (e^{i(t+u)})=i(t+u-2pi)$. Thus when $s=a+ib$, $z_j=r_je^{it_j}$,
begin{align}
(z_1z_2)^s&=exp((a+ib)(log r_1r_2+ i(t_1+t_2-2kpi))\ \
&=r_1^ar_2^a,e^{iblog r_1},e^{iblog r_2},e^{iat_1},e^{iat_2},e^{-bt_1},e^{-bt_2},e^{2bkpi},
end{align}
where $k=0$ if $t_1+t_2leqpi$, and $k=1$ when $t_1+t_2>pi$. On the other hand you have
$$
z_1^sz_2^s=r_1^ar_2^a,e^{iblog r_1},e^{iblog r_2},e^{iat_1},e^{iat_2},e^{-bt_1},e^{-bt_2}.
$$
The two expressions are equal when $t+uleqpi$, and differ by a factor of $e^{2bpi}$ when $t+u>pi$.
In summary, $(z_1z_2)^s=z_1^s+z_2^s$ (in your setup) when $arg z_1+arg z_2leqpi$. This is not terrible, but it prevents you from using the formula in general, when you don't know the arguments of $z_1$ and $z_2$.
answered Dec 17 '18 at 4:51
Martin ArgeramiMartin Argerami
125k1178178
$endgroup$
add a comment |
$begingroup$
The rule $(zw)^s=z^sw^s$ depends on the equality $log(zw)=log z+log w$. But you need the logarithm to be single valued, which requires limiting the argument to one interval, for instance $(-pi,pi]$ as you did; and this generates the following problem. If $t+u>pi$, then $log (e^{i(t+u)})=i(t+u-2pi)$. Thus when $s=a+ib$, $z_j=r_je^{it_j}$,
begin{align}
(z_1z_2)^s&=exp((a+ib)(log r_1r_2+ i(t_1+t_2-2kpi))\ \
&=r_1^ar_2^a,e^{iblog r_1},e^{iblog r_2},e^{iat_1},e^{iat_2},e^{-bt_1},e^{-bt_2},e^{2bkpi},
end{align}
where $k=0$ if $t_1+t_2leqpi$, and $k=1$ when $t_1+t_2>pi$. On the other hand you have
$$
z_1^sz_2^s=r_1^ar_2^a,e^{iblog r_1},e^{iblog r_2},e^{iat_1},e^{iat_2},e^{-bt_1},e^{-bt_2}.
$$
The two expressions are equal when $t+uleqpi$, and differ by a factor of $e^{2bpi}$ when $t+u>pi$.
In summary, $(z_1z_2)^s=z_1^s+z_2^s$ (in your setup) when $arg z_1+arg z_2leqpi$. This is not terrible, but it prevents you from using the formula in general, when you don't know the arguments of $z_1$ and $z_2$.
answered Dec 17 '18 at 4:51
Martin ArgeramiMartin Argerami
125k1178178
$endgroup$
The rule $(zw)^s=z^sw^s$ depends on the equality $log(zw)=log z+log w$. But you need the logarithm to be single valued, which requires limiting the argument to one interval, for instance $(-pi,pi]$ as you did; and this generates the following problem. If $t+u>pi$, then $log (e^{i(t+u)})=i(t+u-2pi)$. Thus when $s=a+ib$, $z_j=r_je^{it_j}$,
begin{align}
(z_1z_2)^s&=exp((a+ib)(log r_1r_2+ i(t_1+t_2-2kpi))\ \
&=r_1^ar_2^a,e^{iblog r_1},e^{iblog r_2},e^{iat_1},e^{iat_2},e^{-bt_1},e^{-bt_2},e^{2bkpi},
end{align}
where $k=0$ if $t_1+t_2leqpi$, and $k=1$ when $t_1+t_2>pi$. On the other hand you have
$$
z_1^sz_2^s=r_1^ar_2^a,e^{iblog r_1},e^{iblog r_2},e^{iat_1},e^{iat_2},e^{-bt_1},e^{-bt_2}.
$$
The two expressions are equal when $t+uleqpi$, and differ by a factor of $e^{2bpi}$ when $t+u>pi$.
In summary, $(z_1z_2)^s=z_1^s+z_2^s$ (in your setup) when $arg z_1+arg z_2leqpi$. This is not terrible, but it prevents you from using the formula in general, when you don't know the arguments of $z_1$ and $z_2$.
answered Dec 17 '18 at 4:51
Martin ArgeramiMartin Argerami
125k1178178
answered Dec 17 '18 at 4:51
Martin ArgeramiMartin Argerami
125k1178178
answered Dec 17 '18 at 4:51
Martin ArgeramiMartin Argerami
125k1178178
answered Dec 17 '18 at 4:51
Martin ArgeramiMartin Argerami
125k1178178
125k1178178
add a comment |
add a comment |
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