Mixing
Complex Roots question.
0
$begingroup$
How many complex solutions does 2^π have? Obviously, something like 2^(7/3) has 3 complex solutions, ie. 3 complex numbers c such that 2∈ c^(3/7). What is the set of c such that 2∈c^(1/π)? Is it infiniteny many solutions, because π is irrational? Can a set be defined containing all solutions?
complex-numbers roots exponentiation
asked Jan 25 at 18:46
Zachary IsaacsonZachary Isaacson
1
$endgroup$
|
show 2 more comments
0
$begingroup$
How many complex solutions does 2^π have? Obviously, something like 2^(7/3) has 3 complex solutions, ie. 3 complex numbers c such that 2∈ c^(3/7). What is the set of c such that 2∈c^(1/π)? Is it infiniteny many solutions, because π is irrational? Can a set be defined containing all solutions?
complex-numbers roots exponentiation
asked Jan 25 at 18:46
Zachary IsaacsonZachary Isaacson
1
$endgroup$
1
$begingroup$
"$2^{pi}$" is not an equation, so it has no solutions; it is a complex number (namely, $exp(pilog(2))$, for some specific branch of $log$. Likewise, "$2^{7/3}$" is a number, not an equation, so it has no solutions. It makes no sense to say "$2in c^{3/7}$", because both sides of that statement are numbers, and numbers are not elements of each other. You need to clarify exactly what it is you are asking.
$endgroup$
– Arturo Magidin
Jan 25 at 18:50
$begingroup$
What if I were to write it like this: {c^(1/π)=2| c ∈C}. I'm saying what is the cardinality of that set, and is there an different way to write such a set?
$endgroup$
– Zachary Isaacson
Jan 25 at 18:57
$begingroup$
If you wrote that, then it would be (i) an attempt at describing a set, but (ii) a failed one, because what you write does not parse. Do you mean: "How many complex numbers $x$ satisfy the equation $x^{1/pi}=2$?" ? If so, you need to be clear about what you mean by the exponentiation, because the usual definition of $a^b$ is $a^b = exp(blog(a))$, but the logarithm function in complex numbers is multivalued.
$endgroup$
– Arturo Magidin
Jan 25 at 18:59
$begingroup$
Ah I see. Well I don't think I know enough math to ask the question properly, or to know of it's even a question I'm able to ask. exp() doesn't mean much to me. My understanding of complex exponentiiation is that it raises the real number radius to the power and finds the real number answer, and perforce the rotation along theta the number of times equal to the power
$endgroup$
– Zachary Isaacson
Jan 25 at 19:05
$begingroup$
That's the exponential function. Just like in the real numbers. What does $3^{pi}$ mean in the real numbers? It means $e^{piln(3)} = exp(piln(3))$. The difference is that $ln$ for real numbers is a single-valued function with limited domain. The equivalent function in the complex numbers is not, because the definition of the exponential is that if $c=a+bi$, with $a,binmathbb{R}$, then $e^c$ is defined to be $e^c = e^a(cos(b)+isin(b))$, so the function is not one-to-one, so the logarithm is multi-valued: there are infinitely many $r$ such that $e^r=c$ for any given $cneq 0$.
$endgroup$
– Arturo Magidin
Jan 25 at 19:09
|
show 2 more comments
0
0
0
$begingroup$
How many complex solutions does 2^π have? Obviously, something like 2^(7/3) has 3 complex solutions, ie. 3 complex numbers c such that 2∈ c^(3/7). What is the set of c such that 2∈c^(1/π)? Is it infiniteny many solutions, because π is irrational? Can a set be defined containing all solutions?
complex-numbers roots exponentiation
asked Jan 25 at 18:46
Zachary IsaacsonZachary Isaacson
1
$endgroup$
How many complex solutions does 2^π have? Obviously, something like 2^(7/3) has 3 complex solutions, ie. 3 complex numbers c such that 2∈ c^(3/7). What is the set of c such that 2∈c^(1/π)? Is it infiniteny many solutions, because π is irrational? Can a set be defined containing all solutions?
complex-numbers roots exponentiation
complex-numbers roots exponentiation
asked Jan 25 at 18:46
Zachary IsaacsonZachary Isaacson
1
asked Jan 25 at 18:46
Zachary IsaacsonZachary Isaacson
1
asked Jan 25 at 18:46
Zachary IsaacsonZachary Isaacson
1
asked Jan 25 at 18:46
Zachary IsaacsonZachary Isaacson
1
asked Jan 25 at 18:46
Zachary IsaacsonZachary Isaacson
1
1
1
$begingroup$
"$2^{pi}$" is not an equation, so it has no solutions; it is a complex number (namely, $exp(pilog(2))$, for some specific branch of $log$. Likewise, "$2^{7/3}$" is a number, not an equation, so it has no solutions. It makes no sense to say "$2in c^{3/7}$", because both sides of that statement are numbers, and numbers are not elements of each other. You need to clarify exactly what it is you are asking.
$endgroup$
– Arturo Magidin
Jan 25 at 18:50
$begingroup$
What if I were to write it like this: {c^(1/π)=2| c ∈C}. I'm saying what is the cardinality of that set, and is there an different way to write such a set?
$endgroup$
– Zachary Isaacson
Jan 25 at 18:57
$begingroup$
If you wrote that, then it would be (i) an attempt at describing a set, but (ii) a failed one, because what you write does not parse. Do you mean: "How many complex numbers $x$ satisfy the equation $x^{1/pi}=2$?" ? If so, you need to be clear about what you mean by the exponentiation, because the usual definition of $a^b$ is $a^b = exp(blog(a))$, but the logarithm function in complex numbers is multivalued.
$endgroup$
– Arturo Magidin
Jan 25 at 18:59
$begingroup$
Ah I see. Well I don't think I know enough math to ask the question properly, or to know of it's even a question I'm able to ask. exp() doesn't mean much to me. My understanding of complex exponentiiation is that it raises the real number radius to the power and finds the real number answer, and perforce the rotation along theta the number of times equal to the power
$endgroup$
– Zachary Isaacson
Jan 25 at 19:05
$begingroup$
That's the exponential function. Just like in the real numbers. What does $3^{pi}$ mean in the real numbers? It means $e^{piln(3)} = exp(piln(3))$. The difference is that $ln$ for real numbers is a single-valued function with limited domain. The equivalent function in the complex numbers is not, because the definition of the exponential is that if $c=a+bi$, with $a,binmathbb{R}$, then $e^c$ is defined to be $e^c = e^a(cos(b)+isin(b))$, so the function is not one-to-one, so the logarithm is multi-valued: there are infinitely many $r$ such that $e^r=c$ for any given $cneq 0$.
$endgroup$
– Arturo Magidin
Jan 25 at 19:09
|
show 2 more comments
1
$begingroup$
"$2^{pi}$" is not an equation, so it has no solutions; it is a complex number (namely, $exp(pilog(2))$, for some specific branch of $log$. Likewise, "$2^{7/3}$" is a number, not an equation, so it has no solutions. It makes no sense to say "$2in c^{3/7}$", because both sides of that statement are numbers, and numbers are not elements of each other. You need to clarify exactly what it is you are asking.
$endgroup$
– Arturo Magidin
Jan 25 at 18:50
$begingroup$
What if I were to write it like this: {c^(1/π)=2| c ∈C}. I'm saying what is the cardinality of that set, and is there an different way to write such a set?
$endgroup$
– Zachary Isaacson
Jan 25 at 18:57
$begingroup$
If you wrote that, then it would be (i) an attempt at describing a set, but (ii) a failed one, because what you write does not parse. Do you mean: "How many complex numbers $x$ satisfy the equation $x^{1/pi}=2$?" ? If so, you need to be clear about what you mean by the exponentiation, because the usual definition of $a^b$ is $a^b = exp(blog(a))$, but the logarithm function in complex numbers is multivalued.
$endgroup$
– Arturo Magidin
Jan 25 at 18:59
$begingroup$
Ah I see. Well I don't think I know enough math to ask the question properly, or to know of it's even a question I'm able to ask. exp() doesn't mean much to me. My understanding of complex exponentiiation is that it raises the real number radius to the power and finds the real number answer, and perforce the rotation along theta the number of times equal to the power
$endgroup$
– Zachary Isaacson
Jan 25 at 19:05
$begingroup$
That's the exponential function. Just like in the real numbers. What does $3^{pi}$ mean in the real numbers? It means $e^{piln(3)} = exp(piln(3))$. The difference is that $ln$ for real numbers is a single-valued function with limited domain. The equivalent function in the complex numbers is not, because the definition of the exponential is that if $c=a+bi$, with $a,binmathbb{R}$, then $e^c$ is defined to be $e^c = e^a(cos(b)+isin(b))$, so the function is not one-to-one, so the logarithm is multi-valued: there are infinitely many $r$ such that $e^r=c$ for any given $cneq 0$.
$endgroup$
– Arturo Magidin
Jan 25 at 19:09
1
1
$begingroup$
"$2^{pi}$" is not an equation, so it has no solutions; it is a complex number (namely, $exp(pilog(2))$, for some specific branch of $log$. Likewise, "$2^{7/3}$" is a number, not an equation, so it has no solutions. It makes no sense to say "$2in c^{3/7}$", because both sides of that statement are numbers, and numbers are not elements of each other. You need to clarify exactly what it is you are asking.
$endgroup$
– Arturo Magidin
Jan 25 at 18:50
$begingroup$
"$2^{pi}$" is not an equation, so it has no solutions; it is a complex number (namely, $exp(pilog(2))$, for some specific branch of $log$. Likewise, "$2^{7/3}$" is a number, not an equation, so it has no solutions. It makes no sense to say "$2in c^{3/7}$", because both sides of that statement are numbers, and numbers are not elements of each other. You need to clarify exactly what it is you are asking.
$endgroup$
– Arturo Magidin
Jan 25 at 18:50
$begingroup$
What if I were to write it like this: {c^(1/π)=2| c ∈C}. I'm saying what is the cardinality of that set, and is there an different way to write such a set?
$endgroup$
– Zachary Isaacson
Jan 25 at 18:57
$begingroup$
What if I were to write it like this: {c^(1/π)=2| c ∈C}. I'm saying what is the cardinality of that set, and is there an different way to write such a set?
$endgroup$
– Zachary Isaacson
Jan 25 at 18:57
$begingroup$
If you wrote that, then it would be (i) an attempt at describing a set, but (ii) a failed one, because what you write does not parse. Do you mean: "How many complex numbers $x$ satisfy the equation $x^{1/pi}=2$?" ? If so, you need to be clear about what you mean by the exponentiation, because the usual definition of $a^b$ is $a^b = exp(blog(a))$, but the logarithm function in complex numbers is multivalued.
$endgroup$
– Arturo Magidin
Jan 25 at 18:59
$begingroup$
If you wrote that, then it would be (i) an attempt at describing a set, but (ii) a failed one, because what you write does not parse. Do you mean: "How many complex numbers $x$ satisfy the equation $x^{1/pi}=2$?" ? If so, you need to be clear about what you mean by the exponentiation, because the usual definition of $a^b$ is $a^b = exp(blog(a))$, but the logarithm function in complex numbers is multivalued.
$endgroup$
– Arturo Magidin
Jan 25 at 18:59
$begingroup$
Ah I see. Well I don't think I know enough math to ask the question properly, or to know of it's even a question I'm able to ask. exp() doesn't mean much to me. My understanding of complex exponentiiation is that it raises the real number radius to the power and finds the real number answer, and perforce the rotation along theta the number of times equal to the power
$endgroup$
– Zachary Isaacson
Jan 25 at 19:05
$begingroup$
Ah I see. Well I don't think I know enough math to ask the question properly, or to know of it's even a question I'm able to ask. exp() doesn't mean much to me. My understanding of complex exponentiiation is that it raises the real number radius to the power and finds the real number answer, and perforce the rotation along theta the number of times equal to the power
$endgroup$
– Zachary Isaacson
Jan 25 at 19:05
$begingroup$
That's the exponential function. Just like in the real numbers. What does $3^{pi}$ mean in the real numbers? It means $e^{piln(3)} = exp(piln(3))$. The difference is that $ln$ for real numbers is a single-valued function with limited domain. The equivalent function in the complex numbers is not, because the definition of the exponential is that if $c=a+bi$, with $a,binmathbb{R}$, then $e^c$ is defined to be $e^c = e^a(cos(b)+isin(b))$, so the function is not one-to-one, so the logarithm is multi-valued: there are infinitely many $r$ such that $e^r=c$ for any given $cneq 0$.
$endgroup$
– Arturo Magidin
Jan 25 at 19:09
$begingroup$
That's the exponential function. Just like in the real numbers. What does $3^{pi}$ mean in the real numbers? It means $e^{piln(3)} = exp(piln(3))$. The difference is that $ln$ for real numbers is a single-valued function with limited domain. The equivalent function in the complex numbers is not, because the definition of the exponential is that if $c=a+bi$, with $a,binmathbb{R}$, then $e^c$ is defined to be $e^c = e^a(cos(b)+isin(b))$, so the function is not one-to-one, so the logarithm is multi-valued: there are infinitely many $r$ such that $e^r=c$ for any given $cneq 0$.
$endgroup$
– Arturo Magidin
Jan 25 at 19:09
|
show 2 more comments
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1
$begingroup$
"$2^{pi}$" is not an equation, so it has no solutions; it is a complex number (namely, $exp(pilog(2))$, for some specific branch of $log$. Likewise, "$2^{7/3}$" is a number, not an equation, so it has no solutions. It makes no sense to say "$2in c^{3/7}$", because both sides of that statement are numbers, and numbers are not elements of each other. You need to clarify exactly what it is you are asking.
$endgroup$
– Arturo Magidin
Jan 25 at 18:50
$begingroup$
What if I were to write it like this: {c^(1/π)=2| c ∈C}. I'm saying what is the cardinality of that set, and is there an different way to write such a set?
$endgroup$
– Zachary Isaacson
Jan 25 at 18:57
$begingroup$
If you wrote that, then it would be (i) an attempt at describing a set, but (ii) a failed one, because what you write does not parse. Do you mean: "How many complex numbers $x$ satisfy the equation $x^{1/pi}=2$?" ? If so, you need to be clear about what you mean by the exponentiation, because the usual definition of $a^b$ is $a^b = exp(blog(a))$, but the logarithm function in complex numbers is multivalued.
$endgroup$
– Arturo Magidin
Jan 25 at 18:59
$begingroup$
Ah I see. Well I don't think I know enough math to ask the question properly, or to know of it's even a question I'm able to ask. exp() doesn't mean much to me. My understanding of complex exponentiiation is that it raises the real number radius to the power and finds the real number answer, and perforce the rotation along theta the number of times equal to the power
$endgroup$
– Zachary Isaacson
Jan 25 at 19:05
$begingroup$
That's the exponential function. Just like in the real numbers. What does $3^{pi}$ mean in the real numbers? It means $e^{piln(3)} = exp(piln(3))$. The difference is that $ln$ for real numbers is a single-valued function with limited domain. The equivalent function in the complex numbers is not, because the definition of the exponential is that if $c=a+bi$, with $a,binmathbb{R}$, then $e^c$ is defined to be $e^c = e^a(cos(b)+isin(b))$, so the function is not one-to-one, so the logarithm is multi-valued: there are infinitely many $r$ such that $e^r=c$ for any given $cneq 0$.
$endgroup$
– Arturo Magidin
Jan 25 at 19:09