Mixing
Find the different principal values. [closed]
Show that the principal value of $i^{i^i}$ differs from the principal value of $i^{i*i}$.
And find the set of all values of the expression $i^{i^i}$.
complex-analysis
asked Nov 21 '18 at 12:27
Peter van de Berg
198
closed as off-topic by Did, Adrian Keister, supinf, Paul Plummer, amWhy Nov 29 '18 at 17:00
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Show that the principal value of $i^{i^i}$ differs from the principal value of $i^{i*i}$.
And find the set of all values of the expression $i^{i^i}$.
complex-analysis
asked Nov 21 '18 at 12:27
Peter van de Berg
198
closed as off-topic by Did, Adrian Keister, supinf, Paul Plummer, amWhy Nov 29 '18 at 17:00
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Did, Adrian Keister, supinf, Paul Plummer, amWhy
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
Show that the principal value of $i^{i^i}$ differs from the principal value of $i^{i*i}$.
And find the set of all values of the expression $i^{i^i}$.
complex-analysis
asked Nov 21 '18 at 12:27
Peter van de Berg
198
Show that the principal value of $i^{i^i}$ differs from the principal value of $i^{i*i}$.
And find the set of all values of the expression $i^{i^i}$.
complex-analysis
complex-analysis
asked Nov 21 '18 at 12:27
Peter van de Berg
198
asked Nov 21 '18 at 12:27
Peter van de Berg
198
asked Nov 21 '18 at 12:27
Peter van de Berg
198
asked Nov 21 '18 at 12:27
Peter van de Berg
198
asked Nov 21 '18 at 12:27
Peter van de Berg
198
198
closed as off-topic by Did, Adrian Keister, supinf, Paul Plummer, amWhy Nov 29 '18 at 17:00
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Did, Adrian Keister, supinf, Paul Plummer, amWhy
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Did, Adrian Keister, supinf, Paul Plummer, amWhy Nov 29 '18 at 17:00
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Did, Adrian Keister, supinf, Paul Plummer, amWhy
If this question can be reworded to fit the rules in the help center, please edit the question.
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1 Answer
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First, $i^{i*i}=i^{-1}=frac{1}{i}=-i$.
Then, $i^{i^i}=i^{(e^{frac{pi}{2}i+2pi mi})^i}=i^{e^{-frac{pi}{2}-2pi m}}=(cis(frac{pi}{2}))^{(e^{-frac{pi}{2}-2pi m})}=cis(frac{pi}{2}e^{-frac{pi}{2}-2pi m})$, using the notation $cis(x)=cos(x)+i*sin(x)$, and for some integer $m$. The principal value will be for $m=0$, so $i^{i^i}=cis(frac{pi}{2}e^{-frac{pi}{2}})$.
In order to show that there is no value for $m$ for which the two expressions are equal, we will suppose that the value $m=n$ will fulfill $cis(frac{pi}{2}e^{-frac{pi}{2}-2pi m})=-i$
$$-i=cis(frac{3pi}{2})=cis(frac{pi}{2}e^{-frac{pi}{2}-2pi n})$$
$$frac{3pi}{2}=frac{pi}{2} e^{-frac{pi}{2}-2pi n} $$
$$3=e^{-frac{pi}{2}-2pi n}$$
$$frac{1}{3}=e^{frac{pi}{2}+2pi n}$$
$$n=-frac{frac{pi}{2}+ln(3)}{2pi}=-frac{1}{4}-frac{ln(3)}{2pi}$$
As $frac{ln(3)}{2pi}$ is irrational, $n$ must be irrational, and as such, cannot be equal to any integer $m$.
answered Nov 21 '18 at 13:08
MoKo19
1914
Thankyou so much! This will help me
– Peter van de Berg
Nov 22 '18 at 12:21
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
First, $i^{i*i}=i^{-1}=frac{1}{i}=-i$.
Then, $i^{i^i}=i^{(e^{frac{pi}{2}i+2pi mi})^i}=i^{e^{-frac{pi}{2}-2pi m}}=(cis(frac{pi}{2}))^{(e^{-frac{pi}{2}-2pi m})}=cis(frac{pi}{2}e^{-frac{pi}{2}-2pi m})$, using the notation $cis(x)=cos(x)+i*sin(x)$, and for some integer $m$. The principal value will be for $m=0$, so $i^{i^i}=cis(frac{pi}{2}e^{-frac{pi}{2}})$.
In order to show that there is no value for $m$ for which the two expressions are equal, we will suppose that the value $m=n$ will fulfill $cis(frac{pi}{2}e^{-frac{pi}{2}-2pi m})=-i$
$$-i=cis(frac{3pi}{2})=cis(frac{pi}{2}e^{-frac{pi}{2}-2pi n})$$
$$frac{3pi}{2}=frac{pi}{2} e^{-frac{pi}{2}-2pi n} $$
$$3=e^{-frac{pi}{2}-2pi n}$$
$$frac{1}{3}=e^{frac{pi}{2}+2pi n}$$
$$n=-frac{frac{pi}{2}+ln(3)}{2pi}=-frac{1}{4}-frac{ln(3)}{2pi}$$
As $frac{ln(3)}{2pi}$ is irrational, $n$ must be irrational, and as such, cannot be equal to any integer $m$.
answered Nov 21 '18 at 13:08
MoKo19
1914
Thankyou so much! This will help me
– Peter van de Berg
Nov 22 '18 at 12:21
add a comment |
First, $i^{i*i}=i^{-1}=frac{1}{i}=-i$.
Then, $i^{i^i}=i^{(e^{frac{pi}{2}i+2pi mi})^i}=i^{e^{-frac{pi}{2}-2pi m}}=(cis(frac{pi}{2}))^{(e^{-frac{pi}{2}-2pi m})}=cis(frac{pi}{2}e^{-frac{pi}{2}-2pi m})$, using the notation $cis(x)=cos(x)+i*sin(x)$, and for some integer $m$. The principal value will be for $m=0$, so $i^{i^i}=cis(frac{pi}{2}e^{-frac{pi}{2}})$.
In order to show that there is no value for $m$ for which the two expressions are equal, we will suppose that the value $m=n$ will fulfill $cis(frac{pi}{2}e^{-frac{pi}{2}-2pi m})=-i$
$$-i=cis(frac{3pi}{2})=cis(frac{pi}{2}e^{-frac{pi}{2}-2pi n})$$
$$frac{3pi}{2}=frac{pi}{2} e^{-frac{pi}{2}-2pi n} $$
$$3=e^{-frac{pi}{2}-2pi n}$$
$$frac{1}{3}=e^{frac{pi}{2}+2pi n}$$
$$n=-frac{frac{pi}{2}+ln(3)}{2pi}=-frac{1}{4}-frac{ln(3)}{2pi}$$
As $frac{ln(3)}{2pi}$ is irrational, $n$ must be irrational, and as such, cannot be equal to any integer $m$.
answered Nov 21 '18 at 13:08
MoKo19
1914
Thankyou so much! This will help me
– Peter van de Berg
Nov 22 '18 at 12:21
add a comment |
First, $i^{i*i}=i^{-1}=frac{1}{i}=-i$.
Then, $i^{i^i}=i^{(e^{frac{pi}{2}i+2pi mi})^i}=i^{e^{-frac{pi}{2}-2pi m}}=(cis(frac{pi}{2}))^{(e^{-frac{pi}{2}-2pi m})}=cis(frac{pi}{2}e^{-frac{pi}{2}-2pi m})$, using the notation $cis(x)=cos(x)+i*sin(x)$, and for some integer $m$. The principal value will be for $m=0$, so $i^{i^i}=cis(frac{pi}{2}e^{-frac{pi}{2}})$.
In order to show that there is no value for $m$ for which the two expressions are equal, we will suppose that the value $m=n$ will fulfill $cis(frac{pi}{2}e^{-frac{pi}{2}-2pi m})=-i$
$$-i=cis(frac{3pi}{2})=cis(frac{pi}{2}e^{-frac{pi}{2}-2pi n})$$
$$frac{3pi}{2}=frac{pi}{2} e^{-frac{pi}{2}-2pi n} $$
$$3=e^{-frac{pi}{2}-2pi n}$$
$$frac{1}{3}=e^{frac{pi}{2}+2pi n}$$
$$n=-frac{frac{pi}{2}+ln(3)}{2pi}=-frac{1}{4}-frac{ln(3)}{2pi}$$
As $frac{ln(3)}{2pi}$ is irrational, $n$ must be irrational, and as such, cannot be equal to any integer $m$.
answered Nov 21 '18 at 13:08
MoKo19
1914
First, $i^{i*i}=i^{-1}=frac{1}{i}=-i$.
Then, $i^{i^i}=i^{(e^{frac{pi}{2}i+2pi mi})^i}=i^{e^{-frac{pi}{2}-2pi m}}=(cis(frac{pi}{2}))^{(e^{-frac{pi}{2}-2pi m})}=cis(frac{pi}{2}e^{-frac{pi}{2}-2pi m})$, using the notation $cis(x)=cos(x)+i*sin(x)$, and for some integer $m$. The principal value will be for $m=0$, so $i^{i^i}=cis(frac{pi}{2}e^{-frac{pi}{2}})$.
In order to show that there is no value for $m$ for which the two expressions are equal, we will suppose that the value $m=n$ will fulfill $cis(frac{pi}{2}e^{-frac{pi}{2}-2pi m})=-i$
$$-i=cis(frac{3pi}{2})=cis(frac{pi}{2}e^{-frac{pi}{2}-2pi n})$$
$$frac{3pi}{2}=frac{pi}{2} e^{-frac{pi}{2}-2pi n} $$
$$3=e^{-frac{pi}{2}-2pi n}$$
$$frac{1}{3}=e^{frac{pi}{2}+2pi n}$$
$$n=-frac{frac{pi}{2}+ln(3)}{2pi}=-frac{1}{4}-frac{ln(3)}{2pi}$$
As $frac{ln(3)}{2pi}$ is irrational, $n$ must be irrational, and as such, cannot be equal to any integer $m$.
answered Nov 21 '18 at 13:08
MoKo19
1914
answered Nov 21 '18 at 13:08
MoKo19
1914
answered Nov 21 '18 at 13:08
MoKo19
1914
answered Nov 21 '18 at 13:08
MoKo19
1914
1914
Thankyou so much! This will help me
– Peter van de Berg
Nov 22 '18 at 12:21
add a comment |
Thankyou so much! This will help me
– Peter van de Berg
Nov 22 '18 at 12:21
Thankyou so much! This will help me
– Peter van de Berg
Nov 22 '18 at 12:21
Thankyou so much! This will help me
– Peter van de Berg
Nov 22 '18 at 12:21
add a comment |
