How to convert from complex number to polar coordinates (solve for r and theta).

1

$begingroup$

How can I convert $ze^z$ to polar coordinates and what are the answers for ($r$ and $theta$)?

I've been trying to solve the question for so long, but I unfortunately didn't succeed.

complex-numbers

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edited Jan 23 at 16:02

saulspatz

17.3k31435

asked Jan 23 at 15:57

shadishadi

61

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Is $z$ real or complex?
  $endgroup$
  – Henry
  Jan 23 at 16:03
  
  
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Please read the tag descriptions before assigning them. This question has nothing to do with algebraic geometry, which is a very advanced subject.
  $endgroup$
  – saulspatz
  Jan 23 at 16:04
  

  
  
  
  

add a comment | 

1

$begingroup$

How can I convert $ze^z$ to polar coordinates and what are the answers for ($r$ and $theta$)?

I've been trying to solve the question for so long, but I unfortunately didn't succeed.

complex-numbers

share|cite|improve this question

edited Jan 23 at 16:02

saulspatz

17.3k31435

asked Jan 23 at 15:57

shadishadi

61

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Is $z$ real or complex?
  $endgroup$
  – Henry
  Jan 23 at 16:03
  
  
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Please read the tag descriptions before assigning them. This question has nothing to do with algebraic geometry, which is a very advanced subject.
  $endgroup$
  – saulspatz
  Jan 23 at 16:04
  

  
  
  
  

add a comment | 

1

1

1

$begingroup$

How can I convert $ze^z$ to polar coordinates and what are the answers for ($r$ and $theta$)?

I've been trying to solve the question for so long, but I unfortunately didn't succeed.

complex-numbers

share|cite|improve this question

edited Jan 23 at 16:02

saulspatz

17.3k31435

asked Jan 23 at 15:57

shadishadi

61

$endgroup$

How can I convert $ze^z$ to polar coordinates and what are the answers for ($r$ and $theta$)?

I've been trying to solve the question for so long, but I unfortunately didn't succeed.

complex-numbers

complex-numbers

share|cite|improve this question

edited Jan 23 at 16:02

saulspatz

17.3k31435

asked Jan 23 at 15:57

shadishadi

61

share|cite|improve this question

edited Jan 23 at 16:02

saulspatz

17.3k31435

asked Jan 23 at 15:57

shadishadi

61

share|cite|improve this question

share|cite|improve this question

edited Jan 23 at 16:02

saulspatz

17.3k31435

edited Jan 23 at 16:02

saulspatz

17.3k31435

edited Jan 23 at 16:02

saulspatz

17.3k31435

17.3k31435

asked Jan 23 at 15:57

shadishadi

61

asked Jan 23 at 15:57

shadishadi

61

asked Jan 23 at 15:57

shadishadi

61

61

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Is $z$ real or complex?
  $endgroup$
  – Henry
  Jan 23 at 16:03
  
  
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Please read the tag descriptions before assigning them. This question has nothing to do with algebraic geometry, which is a very advanced subject.
  $endgroup$
  – saulspatz
  Jan 23 at 16:04
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Is $z$ real or complex?
  $endgroup$
  – Henry
  Jan 23 at 16:03
  
  
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Please read the tag descriptions before assigning them. This question has nothing to do with algebraic geometry, which is a very advanced subject.
  $endgroup$
  – saulspatz
  Jan 23 at 16:04
  

  
  
  
  

$begingroup$
Is $z$ real or complex?
$endgroup$
– Henry
Jan 23 at 16:03

$begingroup$
Is $z$ real or complex?
$endgroup$
– Henry
Jan 23 at 16:03

2

2

$begingroup$
Please read the tag descriptions before assigning them. This question has nothing to do with algebraic geometry, which is a very advanced subject.
$endgroup$
– saulspatz
Jan 23 at 16:04

$begingroup$
Please read the tag descriptions before assigning them. This question has nothing to do with algebraic geometry, which is a very advanced subject.
$endgroup$
– saulspatz
Jan 23 at 16:04

add a comment | 

2 Answers
2

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1

$begingroup$

Let $z = p e^{ipsi}$, where $|z| = p$ and $arg(z) = psi$. This implies that,$$
begin{align} z e^z &= p e^{i psi} e^{p e^{i psi}} \ &= p e^{i psi}e^{pcos psi + i p sin psi} \ &= p e^{i psi}e^{pcos psi} e^{ i p sin psi} \&= p e^{pcos psi} e^{i (psi + p sin psi)}
end{align}$$
Thus, we have written $z e^z$ in the polar form where $r = p e^{pcos psi}$ and $theta = psi + p sin psi$.

share|cite|improve this answer

answered Jan 23 at 16:20

Ajay Kumar NairAjay Kumar Nair

1109

$endgroup$

add a comment | 

1

$begingroup$

Let $x=ze^z$, $z=re^{(itheta)}=r(costheta+isintheta)$.

So, $x=re^{itheta}e^{r(costheta+isintheta)}=re^{itheta}e^{rcostheta+irsintheta}$

Therefore, $x=re^{rcostheta}e^{i(theta+sintheta)}$

On solving, $$x=re^{rcostheta}(cos(theta+sintheta)+isin(theta+sintheta))$$

Or, $$x=|z|e^{Re(z)}(cos(theta+sintheta)+isin(theta+sintheta))$$

Or, $$x=|z|e^{Re(z)}Bigg(cosbigg(theta+frac{Im(z)}{r}bigg)+isinbigg(theta+frac{Im(z)}{r}bigg)Bigg)$$
The previous three equations are same, consider whichever suffices your usage.

share|cite|improve this answer

edited Jan 23 at 16:38

answered Jan 23 at 16:19

Mayank M.Mayank M.

493413

$endgroup$

add a comment | 

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2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

active

oldest

votes

active

oldest

votes

1

$begingroup$

Let $z = p e^{ipsi}$, where $|z| = p$ and $arg(z) = psi$. This implies that,$$
begin{align} z e^z &= p e^{i psi} e^{p e^{i psi}} \ &= p e^{i psi}e^{pcos psi + i p sin psi} \ &= p e^{i psi}e^{pcos psi} e^{ i p sin psi} \&= p e^{pcos psi} e^{i (psi + p sin psi)}
end{align}$$
Thus, we have written $z e^z$ in the polar form where $r = p e^{pcos psi}$ and $theta = psi + p sin psi$.

share|cite|improve this answer

answered Jan 23 at 16:20

Ajay Kumar NairAjay Kumar Nair

1109

$endgroup$

add a comment | 

1

$begingroup$

Let $z = p e^{ipsi}$, where $|z| = p$ and $arg(z) = psi$. This implies that,$$
begin{align} z e^z &= p e^{i psi} e^{p e^{i psi}} \ &= p e^{i psi}e^{pcos psi + i p sin psi} \ &= p e^{i psi}e^{pcos psi} e^{ i p sin psi} \&= p e^{pcos psi} e^{i (psi + p sin psi)}
end{align}$$
Thus, we have written $z e^z$ in the polar form where $r = p e^{pcos psi}$ and $theta = psi + p sin psi$.

share|cite|improve this answer

answered Jan 23 at 16:20

Ajay Kumar NairAjay Kumar Nair

1109

$endgroup$

add a comment | 

1

1

1

$begingroup$

Let $z = p e^{ipsi}$, where $|z| = p$ and $arg(z) = psi$. This implies that,$$
begin{align} z e^z &= p e^{i psi} e^{p e^{i psi}} \ &= p e^{i psi}e^{pcos psi + i p sin psi} \ &= p e^{i psi}e^{pcos psi} e^{ i p sin psi} \&= p e^{pcos psi} e^{i (psi + p sin psi)}
end{align}$$
Thus, we have written $z e^z$ in the polar form where $r = p e^{pcos psi}$ and $theta = psi + p sin psi$.

share|cite|improve this answer

answered Jan 23 at 16:20

Ajay Kumar NairAjay Kumar Nair

1109

$endgroup$

Let $z = p e^{ipsi}$, where $|z| = p$ and $arg(z) = psi$. This implies that,$$
begin{align} z e^z &= p e^{i psi} e^{p e^{i psi}} \ &= p e^{i psi}e^{pcos psi + i p sin psi} \ &= p e^{i psi}e^{pcos psi} e^{ i p sin psi} \&= p e^{pcos psi} e^{i (psi + p sin psi)}
end{align}$$
Thus, we have written $z e^z$ in the polar form where $r = p e^{pcos psi}$ and $theta = psi + p sin psi$.

share|cite|improve this answer

answered Jan 23 at 16:20

Ajay Kumar NairAjay Kumar Nair

1109

share|cite|improve this answer

share|cite|improve this answer

answered Jan 23 at 16:20

Ajay Kumar NairAjay Kumar Nair

1109

answered Jan 23 at 16:20

Ajay Kumar NairAjay Kumar Nair

1109

answered Jan 23 at 16:20

Ajay Kumar NairAjay Kumar Nair

1109

1109

add a comment | 

add a comment | 

1

$begingroup$

Let $x=ze^z$, $z=re^{(itheta)}=r(costheta+isintheta)$.

So, $x=re^{itheta}e^{r(costheta+isintheta)}=re^{itheta}e^{rcostheta+irsintheta}$

Therefore, $x=re^{rcostheta}e^{i(theta+sintheta)}$

On solving, $$x=re^{rcostheta}(cos(theta+sintheta)+isin(theta+sintheta))$$

Or, $$x=|z|e^{Re(z)}(cos(theta+sintheta)+isin(theta+sintheta))$$

Or, $$x=|z|e^{Re(z)}Bigg(cosbigg(theta+frac{Im(z)}{r}bigg)+isinbigg(theta+frac{Im(z)}{r}bigg)Bigg)$$
The previous three equations are same, consider whichever suffices your usage.

share|cite|improve this answer

edited Jan 23 at 16:38

answered Jan 23 at 16:19

Mayank M.Mayank M.

493413

$endgroup$

add a comment | 

1

$begingroup$

Let $x=ze^z$, $z=re^{(itheta)}=r(costheta+isintheta)$.

So, $x=re^{itheta}e^{r(costheta+isintheta)}=re^{itheta}e^{rcostheta+irsintheta}$

Therefore, $x=re^{rcostheta}e^{i(theta+sintheta)}$

On solving, $$x=re^{rcostheta}(cos(theta+sintheta)+isin(theta+sintheta))$$

Or, $$x=|z|e^{Re(z)}(cos(theta+sintheta)+isin(theta+sintheta))$$

Or, $$x=|z|e^{Re(z)}Bigg(cosbigg(theta+frac{Im(z)}{r}bigg)+isinbigg(theta+frac{Im(z)}{r}bigg)Bigg)$$
The previous three equations are same, consider whichever suffices your usage.

share|cite|improve this answer

edited Jan 23 at 16:38

answered Jan 23 at 16:19

Mayank M.Mayank M.

493413

$endgroup$

add a comment | 

1

1

1

$begingroup$

Let $x=ze^z$, $z=re^{(itheta)}=r(costheta+isintheta)$.

So, $x=re^{itheta}e^{r(costheta+isintheta)}=re^{itheta}e^{rcostheta+irsintheta}$

Therefore, $x=re^{rcostheta}e^{i(theta+sintheta)}$

On solving, $$x=re^{rcostheta}(cos(theta+sintheta)+isin(theta+sintheta))$$

Or, $$x=|z|e^{Re(z)}(cos(theta+sintheta)+isin(theta+sintheta))$$

Or, $$x=|z|e^{Re(z)}Bigg(cosbigg(theta+frac{Im(z)}{r}bigg)+isinbigg(theta+frac{Im(z)}{r}bigg)Bigg)$$
The previous three equations are same, consider whichever suffices your usage.

share|cite|improve this answer

edited Jan 23 at 16:38

answered Jan 23 at 16:19

Mayank M.Mayank M.

493413

$endgroup$

Let $x=ze^z$, $z=re^{(itheta)}=r(costheta+isintheta)$.

So, $x=re^{itheta}e^{r(costheta+isintheta)}=re^{itheta}e^{rcostheta+irsintheta}$

Therefore, $x=re^{rcostheta}e^{i(theta+sintheta)}$

On solving, $$x=re^{rcostheta}(cos(theta+sintheta)+isin(theta+sintheta))$$

Or, $$x=|z|e^{Re(z)}(cos(theta+sintheta)+isin(theta+sintheta))$$

Or, $$x=|z|e^{Re(z)}Bigg(cosbigg(theta+frac{Im(z)}{r}bigg)+isinbigg(theta+frac{Im(z)}{r}bigg)Bigg)$$
The previous three equations are same, consider whichever suffices your usage.

share|cite|improve this answer

edited Jan 23 at 16:38

answered Jan 23 at 16:19

Mayank M.Mayank M.

493413

share|cite|improve this answer

share|cite|improve this answer

edited Jan 23 at 16:38

edited Jan 23 at 16:38

edited Jan 23 at 16:38

answered Jan 23 at 16:19

Mayank M.Mayank M.

493413

answered Jan 23 at 16:19

Mayank M.Mayank M.

493413

answered Jan 23 at 16:19

Mayank M.Mayank M.

493413

493413

add a comment | 

add a comment | 

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Use MathJax to format equations. MathJax reference.

To learn more, see our tips on writing great answers.

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