Find Two Closest Solutions

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Find the two closest solutions to $z=0$ of
$$e^e^z=1$$

I started with setting $t=e^z$

$$e^t=1iff e^t=e^(0+2pi k)iiff t=2pi ki$$

so

Because we are on the unit circle every two solution $|f(z_1)|=|f(z_2)|=1$ so we need to take the square root.

$$e^z=frac2pi k i2 text Where k=0,1$$

$$z=ln|frac2pi k2|=ln|pi k|$$ for $k=1$ we get $z=ln|pi|$ when $k=0$ we get $ln|0|$ so we need to take the 3rd root

$z=ln|frac2pi k3|$ for $k=0,1,2$

for $k=1$ we get $ln|fracpi3|$ for $k=1$ we get $ln|frac2pi3| $

Is it correct?

complex-analysis

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asked Jul 15 at 7:37

newhere

759310

add a comment | 

up vote
0
down vote

favorite

Find the two closest solutions to $z=0$ of
$$e^e^z=1$$

I started with setting $t=e^z$

$$e^t=1iff e^t=e^(0+2pi k)iiff t=2pi ki$$

so

Because we are on the unit circle every two solution $|f(z_1)|=|f(z_2)|=1$ so we need to take the square root.

$$e^z=frac2pi k i2 text Where k=0,1$$

$$z=ln|frac2pi k2|=ln|pi k|$$ for $k=1$ we get $z=ln|pi|$ when $k=0$ we get $ln|0|$ so we need to take the 3rd root

$z=ln|frac2pi k3|$ for $k=0,1,2$

for $k=1$ we get $ln|fracpi3|$ for $k=1$ we get $ln|frac2pi3| $

Is it correct?

complex-analysis

share|cite|improve this question

asked Jul 15 at 7:37

newhere

759310

add a comment | 

up vote
0
down vote

favorite

up vote
0
down vote

favorite

Find the two closest solutions to $z=0$ of
$$e^e^z=1$$

I started with setting $t=e^z$

$$e^t=1iff e^t=e^(0+2pi k)iiff t=2pi ki$$

so

Because we are on the unit circle every two solution $|f(z_1)|=|f(z_2)|=1$ so we need to take the square root.

$$e^z=frac2pi k i2 text Where k=0,1$$

$$z=ln|frac2pi k2|=ln|pi k|$$ for $k=1$ we get $z=ln|pi|$ when $k=0$ we get $ln|0|$ so we need to take the 3rd root

$z=ln|frac2pi k3|$ for $k=0,1,2$

for $k=1$ we get $ln|fracpi3|$ for $k=1$ we get $ln|frac2pi3| $

Is it correct?

complex-analysis

share|cite|improve this question

asked Jul 15 at 7:37

newhere

759310

Find the two closest solutions to $z=0$ of
$$e^e^z=1$$

I started with setting $t=e^z$

$$e^t=1iff e^t=e^(0+2pi k)iiff t=2pi ki$$

so

Because we are on the unit circle every two solution $|f(z_1)|=|f(z_2)|=1$ so we need to take the square root.

$$e^z=frac2pi k i2 text Where k=0,1$$

$$z=ln|frac2pi k2|=ln|pi k|$$ for $k=1$ we get $z=ln|pi|$ when $k=0$ we get $ln|0|$ so we need to take the 3rd root

$z=ln|frac2pi k3|$ for $k=0,1,2$

for $k=1$ we get $ln|fracpi3|$ for $k=1$ we get $ln|frac2pi3| $

Is it correct?

complex-analysis

share|cite|improve this question

asked Jul 15 at 7:37

newhere

759310

share|cite|improve this question

share|cite|improve this question

asked Jul 15 at 7:37

newhere

759310

asked Jul 15 at 7:37

newhere

759310

asked Jul 15 at 7:37

newhere

759310

759310

add a comment | 

add a comment | 

1 Answer
1

active

oldest

votes

up vote
1
down vote



accepted

Changing $e^z = t$ we have $e^t = 1$ so $t=2pi i k$ with $k$ integer, i.e. $e^z = 2pi ik$ with $k$ integer. If $k=0$, $e^z=0$ has no solution, so if $kneq 0$
$$
z = ln|2pi ik| + arg(2pi i k)i = ln(2pi |k|) + arg(2pi i k)i
$$
Simplifying
$$
z = begincases
ln(2pi k) + pi i/2 & k >0\
ln(-2pi k) - pi i/2 & k<0
endcases
$$

share|cite|improve this answer

edited Jul 16 at 15:54

answered Jul 15 at 9:43

Rafael Gonzalez Lopez

652112

• 
  
  
  

  
  

  
  
  
  
  
  

  
  How did we get $pm fracpi i2$?
  – newhere
  Jul 15 at 10:52
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @newhere If $k>0$ then $arg(2pi i k) = pi/2$ (is a pure imaginary number with positive imaginary part). If $k<0$ then $arg(2pi i k) = -pi /2$.
  – Rafael Gonzalez Lopez
  Jul 15 at 17:01
  
  

  
  
  
  

add a comment | 

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1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
1
down vote



accepted

Changing $e^z = t$ we have $e^t = 1$ so $t=2pi i k$ with $k$ integer, i.e. $e^z = 2pi ik$ with $k$ integer. If $k=0$, $e^z=0$ has no solution, so if $kneq 0$
$$
z = ln|2pi ik| + arg(2pi i k)i = ln(2pi |k|) + arg(2pi i k)i
$$
Simplifying
$$
z = begincases
ln(2pi k) + pi i/2 & k >0\
ln(-2pi k) - pi i/2 & k<0
endcases
$$

share|cite|improve this answer

edited Jul 16 at 15:54

answered Jul 15 at 9:43

Rafael Gonzalez Lopez

652112

• 
  
  
  

  
  

  
  
  
  
  
  

  
  How did we get $pm fracpi i2$?
  – newhere
  Jul 15 at 10:52
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @newhere If $k>0$ then $arg(2pi i k) = pi/2$ (is a pure imaginary number with positive imaginary part). If $k<0$ then $arg(2pi i k) = -pi /2$.
  – Rafael Gonzalez Lopez
  Jul 15 at 17:01
  
  

  
  
  
  

add a comment | 

up vote
1
down vote



accepted

Changing $e^z = t$ we have $e^t = 1$ so $t=2pi i k$ with $k$ integer, i.e. $e^z = 2pi ik$ with $k$ integer. If $k=0$, $e^z=0$ has no solution, so if $kneq 0$
$$
z = ln|2pi ik| + arg(2pi i k)i = ln(2pi |k|) + arg(2pi i k)i
$$
Simplifying
$$
z = begincases
ln(2pi k) + pi i/2 & k >0\
ln(-2pi k) - pi i/2 & k<0
endcases
$$

share|cite|improve this answer

edited Jul 16 at 15:54

answered Jul 15 at 9:43

Rafael Gonzalez Lopez

652112

• 
  
  
  

  
  

  
  
  
  
  
  

  
  How did we get $pm fracpi i2$?
  – newhere
  Jul 15 at 10:52
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @newhere If $k>0$ then $arg(2pi i k) = pi/2$ (is a pure imaginary number with positive imaginary part). If $k<0$ then $arg(2pi i k) = -pi /2$.
  – Rafael Gonzalez Lopez
  Jul 15 at 17:01
  
  

  
  
  
  

add a comment | 

up vote
1
down vote



accepted

up vote
1
down vote



accepted

Changing $e^z = t$ we have $e^t = 1$ so $t=2pi i k$ with $k$ integer, i.e. $e^z = 2pi ik$ with $k$ integer. If $k=0$, $e^z=0$ has no solution, so if $kneq 0$
$$
z = ln|2pi ik| + arg(2pi i k)i = ln(2pi |k|) + arg(2pi i k)i
$$
Simplifying
$$
z = begincases
ln(2pi k) + pi i/2 & k >0\
ln(-2pi k) - pi i/2 & k<0
endcases
$$

share|cite|improve this answer

edited Jul 16 at 15:54

answered Jul 15 at 9:43

Rafael Gonzalez Lopez

652112

Changing $e^z = t$ we have $e^t = 1$ so $t=2pi i k$ with $k$ integer, i.e. $e^z = 2pi ik$ with $k$ integer. If $k=0$, $e^z=0$ has no solution, so if $kneq 0$
$$
z = ln|2pi ik| + arg(2pi i k)i = ln(2pi |k|) + arg(2pi i k)i
$$
Simplifying
$$
z = begincases
ln(2pi k) + pi i/2 & k >0\
ln(-2pi k) - pi i/2 & k<0
endcases
$$

share|cite|improve this answer

edited Jul 16 at 15:54

answered Jul 15 at 9:43

Rafael Gonzalez Lopez

652112

share|cite|improve this answer

share|cite|improve this answer

edited Jul 16 at 15:54

edited Jul 16 at 15:54

edited Jul 16 at 15:54

answered Jul 15 at 9:43

Rafael Gonzalez Lopez

652112

answered Jul 15 at 9:43

Rafael Gonzalez Lopez

652112

answered Jul 15 at 9:43

Rafael Gonzalez Lopez

652112

652112

• 
  
  
  

  
  

  
  
  
  
  
  

  
  How did we get $pm fracpi i2$?
  – newhere
  Jul 15 at 10:52
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @newhere If $k>0$ then $arg(2pi i k) = pi/2$ (is a pure imaginary number with positive imaginary part). If $k<0$ then $arg(2pi i k) = -pi /2$.
  – Rafael Gonzalez Lopez
  Jul 15 at 17:01
  
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  How did we get $pm fracpi i2$?
  – newhere
  Jul 15 at 10:52
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @newhere If $k>0$ then $arg(2pi i k) = pi/2$ (is a pure imaginary number with positive imaginary part). If $k<0$ then $arg(2pi i k) = -pi /2$.
  – Rafael Gonzalez Lopez
  Jul 15 at 17:01
  
  

  
  
  
  

How did we get $pm fracpi i2$?
– newhere
Jul 15 at 10:52

How did we get $pm fracpi i2$?
– newhere
Jul 15 at 10:52

1

1

@newhere If $k>0$ then $arg(2pi i k) = pi/2$ (is a pure imaginary number with positive imaginary part). If $k<0$ then $arg(2pi i k) = -pi /2$.
– Rafael Gonzalez Lopez
Jul 15 at 17:01

@newhere If $k>0$ then $arg(2pi i k) = pi/2$ (is a pure imaginary number with positive imaginary part). If $k<0$ then $arg(2pi i k) = -pi /2$.
– Rafael Gonzalez Lopez
Jul 15 at 17:01

add a comment | 

 

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