How to write a complex number in the form $re^i phi$ when $r ge 0$ [closed]

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$$frac1+i1-i$$and$$(1+i)(1-i).$$



The answers are $e^fracipi2$ and $2e^iphi$.






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closed as off-topic by José Carlos Santos, amWhy, Nosrati, Shailesh, Arnaud D. Sep 3 at 10:56


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." – José Carlos Santos, amWhy, Nosrati, Shailesh, Arnaud D.
If this question can be reworded to fit the rules in the help center, please edit the question.












  • To avoid down votes, add your attempts of solving the problem in the question.
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up vote
-3
down vote

favorite












$$frac1+i1-i$$and$$(1+i)(1-i).$$



The answers are $e^fracipi2$ and $2e^iphi$.






complex-numbers







share|cite|improve this question















closed as off-topic by José Carlos Santos, amWhy, Nosrati, Shailesh, Arnaud D. Sep 3 at 10:56


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." – José Carlos Santos, amWhy, Nosrati, Shailesh, Arnaud D.
If this question can be reworded to fit the rules in the help center, please edit the question.












  • To avoid down votes, add your attempts of solving the problem in the question.
    – paulplusx
    Sep 3 at 10:04












up vote
-3
down vote

favorite









up vote
-3
down vote

favorite











$$frac1+i1-i$$and$$(1+i)(1-i).$$



The answers are $e^fracipi2$ and $2e^iphi$.






complex-numbers







share|cite|improve this question















$$frac1+i1-i$$and$$(1+i)(1-i).$$



The answers are $e^fracipi2$ and $2e^iphi$.






complex-numbers




complex-numbers






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Sep 3 at 10:01









José Carlos Santos

122k16101186




122k16101186










asked Sep 3 at 9:39









j.doe

112




112




closed as off-topic by José Carlos Santos, amWhy, Nosrati, Shailesh, Arnaud D. Sep 3 at 10:56


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." – José Carlos Santos, amWhy, Nosrati, Shailesh, Arnaud D.
If this question can be reworded to fit the rules in the help center, please edit the question.




closed as off-topic by José Carlos Santos, amWhy, Nosrati, Shailesh, Arnaud D. Sep 3 at 10:56


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." – José Carlos Santos, amWhy, Nosrati, Shailesh, Arnaud D.
If this question can be reworded to fit the rules in the help center, please edit the question.











  • To avoid down votes, add your attempts of solving the problem in the question.
    – paulplusx
    Sep 3 at 10:04
















  • To avoid down votes, add your attempts of solving the problem in the question.
    – paulplusx
    Sep 3 at 10:04















To avoid down votes, add your attempts of solving the problem in the question.
– paulplusx
Sep 3 at 10:04




To avoid down votes, add your attempts of solving the problem in the question.
– paulplusx
Sep 3 at 10:04










3 Answers
3






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oldest

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up vote
0
down vote



accepted










welcome to MSE! Please, next time try to explain better what is the problem and what you tried to do to solve it, possibly using MathJax.



For the first fraction, first it is better to reduce it:
$$frac1+i1-i=frac(1+i)(1+i)(1+i)(1-i)=frac2i2=i.$$
Then, it is simple to see that the number $r=i$ has radius $1$ and argument $theta=pi/2,$ cause it can be written as $0+1cdot i,$ so $r=sqrt0^2+1^2=1$ and $theta=cotan^-1(0/1)=pi/2.$



For the product, as just said $(i+1)(1-i)=2$ and in this case the radius is equal to $2$ and the argument is $0.$






share|cite|improve this answer
















  • 1




    Ah excellent with this information I was able to understand what to do and figure out my own solution. Thank you.
    – j.doe
    Sep 3 at 10:19


















up vote
1
down vote













Hint:



$$frac1+i1-i=frac(1+i)(1+i)(1-i)(1+i)=frac2i2=i$$



If you plot this point in the complex plane, can you find $phi$ and $r$?



For instance, if you were to plot $z=frac1+i2$ you would get something like:



enter image description here






share|cite|improve this answer





























    up vote
    0
    down vote













    $(1+i)(1-i)=1-i^2=1+1=2$



    So if you are going to write this as $2e^i phi$ then $e^i phi=1$. So $phi=2npi$ where $nin mathbbZ$






    share|cite|improve this answer



























      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes








      up vote
      0
      down vote



      accepted










      welcome to MSE! Please, next time try to explain better what is the problem and what you tried to do to solve it, possibly using MathJax.



      For the first fraction, first it is better to reduce it:
      $$frac1+i1-i=frac(1+i)(1+i)(1+i)(1-i)=frac2i2=i.$$
      Then, it is simple to see that the number $r=i$ has radius $1$ and argument $theta=pi/2,$ cause it can be written as $0+1cdot i,$ so $r=sqrt0^2+1^2=1$ and $theta=cotan^-1(0/1)=pi/2.$



      For the product, as just said $(i+1)(1-i)=2$ and in this case the radius is equal to $2$ and the argument is $0.$






      share|cite|improve this answer
















      • 1




        Ah excellent with this information I was able to understand what to do and figure out my own solution. Thank you.
        – j.doe
        Sep 3 at 10:19















      up vote
      0
      down vote



      accepted










      welcome to MSE! Please, next time try to explain better what is the problem and what you tried to do to solve it, possibly using MathJax.



      For the first fraction, first it is better to reduce it:
      $$frac1+i1-i=frac(1+i)(1+i)(1+i)(1-i)=frac2i2=i.$$
      Then, it is simple to see that the number $r=i$ has radius $1$ and argument $theta=pi/2,$ cause it can be written as $0+1cdot i,$ so $r=sqrt0^2+1^2=1$ and $theta=cotan^-1(0/1)=pi/2.$



      For the product, as just said $(i+1)(1-i)=2$ and in this case the radius is equal to $2$ and the argument is $0.$






      share|cite|improve this answer
















      • 1




        Ah excellent with this information I was able to understand what to do and figure out my own solution. Thank you.
        – j.doe
        Sep 3 at 10:19













      up vote
      0
      down vote



      accepted







      up vote
      0
      down vote



      accepted






      welcome to MSE! Please, next time try to explain better what is the problem and what you tried to do to solve it, possibly using MathJax.



      For the first fraction, first it is better to reduce it:
      $$frac1+i1-i=frac(1+i)(1+i)(1+i)(1-i)=frac2i2=i.$$
      Then, it is simple to see that the number $r=i$ has radius $1$ and argument $theta=pi/2,$ cause it can be written as $0+1cdot i,$ so $r=sqrt0^2+1^2=1$ and $theta=cotan^-1(0/1)=pi/2.$



      For the product, as just said $(i+1)(1-i)=2$ and in this case the radius is equal to $2$ and the argument is $0.$






      share|cite|improve this answer












      welcome to MSE! Please, next time try to explain better what is the problem and what you tried to do to solve it, possibly using MathJax.



      For the first fraction, first it is better to reduce it:
      $$frac1+i1-i=frac(1+i)(1+i)(1+i)(1-i)=frac2i2=i.$$
      Then, it is simple to see that the number $r=i$ has radius $1$ and argument $theta=pi/2,$ cause it can be written as $0+1cdot i,$ so $r=sqrt0^2+1^2=1$ and $theta=cotan^-1(0/1)=pi/2.$



      For the product, as just said $(i+1)(1-i)=2$ and in this case the radius is equal to $2$ and the argument is $0.$







      share|cite|improve this answer












      share|cite|improve this answer



      share|cite|improve this answer










      answered Sep 3 at 9:53









      Riccardo Ceccon

      873320




      873320







      • 1




        Ah excellent with this information I was able to understand what to do and figure out my own solution. Thank you.
        – j.doe
        Sep 3 at 10:19













      • 1




        Ah excellent with this information I was able to understand what to do and figure out my own solution. Thank you.
        – j.doe
        Sep 3 at 10:19








      1




      1




      Ah excellent with this information I was able to understand what to do and figure out my own solution. Thank you.
      – j.doe
      Sep 3 at 10:19





      Ah excellent with this information I was able to understand what to do and figure out my own solution. Thank you.
      – j.doe
      Sep 3 at 10:19











      up vote
      1
      down vote













      Hint:



      $$frac1+i1-i=frac(1+i)(1+i)(1-i)(1+i)=frac2i2=i$$



      If you plot this point in the complex plane, can you find $phi$ and $r$?



      For instance, if you were to plot $z=frac1+i2$ you would get something like:



      enter image description here






      share|cite|improve this answer


























        up vote
        1
        down vote













        Hint:



        $$frac1+i1-i=frac(1+i)(1+i)(1-i)(1+i)=frac2i2=i$$



        If you plot this point in the complex plane, can you find $phi$ and $r$?



        For instance, if you were to plot $z=frac1+i2$ you would get something like:



        enter image description here






        share|cite|improve this answer
























          up vote
          1
          down vote










          up vote
          1
          down vote









          Hint:



          $$frac1+i1-i=frac(1+i)(1+i)(1-i)(1+i)=frac2i2=i$$



          If you plot this point in the complex plane, can you find $phi$ and $r$?



          For instance, if you were to plot $z=frac1+i2$ you would get something like:



          enter image description here






          share|cite|improve this answer














          Hint:



          $$frac1+i1-i=frac(1+i)(1+i)(1-i)(1+i)=frac2i2=i$$



          If you plot this point in the complex plane, can you find $phi$ and $r$?



          For instance, if you were to plot $z=frac1+i2$ you would get something like:



          enter image description here







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Sep 3 at 10:01

























          answered Sep 3 at 9:43









          cansomeonehelpmeout

          5,5383830




          5,5383830




















              up vote
              0
              down vote













              $(1+i)(1-i)=1-i^2=1+1=2$



              So if you are going to write this as $2e^i phi$ then $e^i phi=1$. So $phi=2npi$ where $nin mathbbZ$






              share|cite|improve this answer
























                up vote
                0
                down vote













                $(1+i)(1-i)=1-i^2=1+1=2$



                So if you are going to write this as $2e^i phi$ then $e^i phi=1$. So $phi=2npi$ where $nin mathbbZ$






                share|cite|improve this answer






















                  up vote
                  0
                  down vote










                  up vote
                  0
                  down vote









                  $(1+i)(1-i)=1-i^2=1+1=2$



                  So if you are going to write this as $2e^i phi$ then $e^i phi=1$. So $phi=2npi$ where $nin mathbbZ$






                  share|cite|improve this answer












                  $(1+i)(1-i)=1-i^2=1+1=2$



                  So if you are going to write this as $2e^i phi$ then $e^i phi=1$. So $phi=2npi$ where $nin mathbbZ$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Sep 3 at 9:50









                  Bruce

                  522113




                  522113












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