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Using De Moivre's Theorem $z=3+sqrt2i$ [closed]

De Moivre's Theorem homework.How to find trigonometric values using triangles as fractions?How to find trigonometry values using triangles in 1/2 form?Using De Moivre's theoremSome exact values of $cos theta$ using de Moivre's theoremDe Moivre's theoremUsing the exponential form of a complex number and De Moivre's theoremMaximum value of a trigonometric expression and value of angle giving this max.Using De Moivre's theorem . $(1+i)^100$Using de-Moivre's theorem to find the reciprocal of each number

0

$begingroup$

Given $z=3+sqrt2i$
Find an expression for, $z^5$
Give your answer in the form of $ a+ bi$.

trigonometry

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edited Mar 12 at 20:18

Justin Stevenson

957519

asked Mar 12 at 19:29

Guy CullenGuy Cullen

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Guy Cullen is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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closed as off-topic by Gibbs, RRL, Eevee Trainer, Shailesh, Lee David Chung Lin Mar 13 at 0:35

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 provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Gibbs, RRL, Eevee Trainer, Shailesh, Lee David Chung Lin

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  write your complex number into polar (modulus-argument) form.
  $endgroup$
  – Tito Eliatron
  Mar 12 at 19:30
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  $begingroup$
  Your title says "Using De Moivre's Theorem". Have you tried using that theorem?
  $endgroup$
  – Arthur
  Mar 12 at 19:54
  
  

  
  
  
  

add a comment | 

0

$begingroup$

Given $z=3+sqrt2i$
Find an expression for, $z^5$
Give your answer in the form of $ a+ bi$.

trigonometry

share|cite|improve this question

edited Mar 12 at 20:18

Justin Stevenson

957519

asked Mar 12 at 19:29

Guy CullenGuy Cullen

1

New contributor

Guy Cullen is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

closed as off-topic by Gibbs, RRL, Eevee Trainer, Shailesh, Lee David Chung Lin Mar 13 at 0:35

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 provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Gibbs, RRL, Eevee Trainer, Shailesh, Lee David Chung Lin

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  write your complex number into polar (modulus-argument) form.
  $endgroup$
  – Tito Eliatron
  Mar 12 at 19:30
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  $begingroup$
  Your title says "Using De Moivre's Theorem". Have you tried using that theorem?
  $endgroup$
  – Arthur
  Mar 12 at 19:54
  
  

  
  
  
  

add a comment | 

$begingroup$

Given $z=3+sqrt2i$
Find an expression for, $z^5$
Give your answer in the form of $ a+ bi$.

trigonometry

share|cite|improve this question

edited Mar 12 at 20:18

Justin Stevenson

957519

asked Mar 12 at 19:29

Guy CullenGuy Cullen

1

New contributor

Guy Cullen is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

share|cite|improve this question

edited Mar 12 at 20:18

Justin Stevenson

957519

asked Mar 12 at 19:29

Guy CullenGuy Cullen

1

New contributor

Guy Cullen is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

share|cite|improve this question

edited Mar 12 at 20:18

Justin Stevenson

957519

asked Mar 12 at 19:29

Guy CullenGuy Cullen

1

New contributor

Guy Cullen is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

edited Mar 12 at 20:18

Justin Stevenson

957519

edited Mar 12 at 20:18

Justin Stevenson

957519

asked Mar 12 at 19:29

Guy CullenGuy Cullen

1

New contributor

Guy Cullen is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked Mar 12 at 19:29

Guy CullenGuy Cullen

1

1 Answer
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Using De moivres theorem, you can express $zin mathbbC$ as $r operatorname cis theta$, where r= $sqrtoperatornameRe(z)^2 +operatornameIm(z)^2$. It can also be shown that $(r operatorname cis (theta))^n$$=r^n operatorname cis (ntheta)$ (It follows from the fact that $operatorname cis theta=e^itheta$).

Note that $operatorname cis theta$ $=$ $costheta +i sintheta$.

share|cite|improve this answer

edited Mar 12 at 21:05

YiFan

4,7501727

answered Mar 12 at 20:11

topologicalmagiciantopologicalmagician

1019

$endgroup$

add a comment | 

1

$begingroup$

Using De moivres theorem, you can express $zin mathbbC$ as $r operatorname cis theta$, where r= $sqrtoperatornameRe(z)^2 +operatornameIm(z)^2$. It can also be shown that $(r operatorname cis (theta))^n$$=r^n operatorname cis (ntheta)$ (It follows from the fact that $operatorname cis theta=e^itheta$).

Note that $operatorname cis theta$ $=$ $costheta +i sintheta$.

share|cite|improve this answer

edited Mar 12 at 21:05

YiFan

4,7501727

answered Mar 12 at 20:11

topologicalmagiciantopologicalmagician

1019

$endgroup$

add a comment | 

1

$begingroup$

Using De moivres theorem, you can express $zin mathbbC$ as $r operatorname cis theta$, where r= $sqrtoperatornameRe(z)^2 +operatornameIm(z)^2$. It can also be shown that $(r operatorname cis (theta))^n$$=r^n operatorname cis (ntheta)$ (It follows from the fact that $operatorname cis theta=e^itheta$).

Note that $operatorname cis theta$ $=$ $costheta +i sintheta$.

share|cite|improve this answer

edited Mar 12 at 21:05

YiFan

4,7501727

answered Mar 12 at 20:11

topologicalmagiciantopologicalmagician

1019

$endgroup$

add a comment | 

$begingroup$

Using De moivres theorem, you can express $zin mathbbC$ as $r operatorname cis theta$, where r= $sqrtoperatornameRe(z)^2 +operatornameIm(z)^2$. It can also be shown that $(r operatorname cis (theta))^n$$=r^n operatorname cis (ntheta)$ (It follows from the fact that $operatorname cis theta=e^itheta$).

Note that $operatorname cis theta$ $=$ $costheta +i sintheta$.

share|cite|improve this answer

edited Mar 12 at 21:05

YiFan

4,7501727

answered Mar 12 at 20:11

topologicalmagiciantopologicalmagician

1019

$endgroup$

share|cite|improve this answer

edited Mar 12 at 21:05

YiFan

4,7501727

answered Mar 12 at 20:11

topologicalmagiciantopologicalmagician

1019

edited Mar 12 at 21:05

YiFan

4,7501727

edited Mar 12 at 21:05

YiFan

4,7501727

answered Mar 12 at 20:11

topologicalmagiciantopologicalmagician

1019

answered Mar 12 at 20:11

topologicalmagiciantopologicalmagician

1019