Find the image of the region ${zinmathbb{C}: 0leq Re(z) leq 1/2,; -pi leq Im(z) leq pi}$ under the mapping...











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I have no clue if I am doing this correctly, just basing it off what my teacher did for a similar example.



I look at image of $y=a$



$u(x,y) = e^x cos{a}$



$v(x,y) = e^x sin{a}$



$$y=-pi , f(z) = u+iv = -e^x <0$$
$$y=pi , f(z) = u+iv = -e^x <0$$



So this is below the x-axis, I am confused on how to apply this to the Re(z) boundary if this method is even correct.










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    up vote
    0
    down vote

    favorite












    I have no clue if I am doing this correctly, just basing it off what my teacher did for a similar example.



    I look at image of $y=a$



    $u(x,y) = e^x cos{a}$



    $v(x,y) = e^x sin{a}$



    $$y=-pi , f(z) = u+iv = -e^x <0$$
    $$y=pi , f(z) = u+iv = -e^x <0$$



    So this is below the x-axis, I am confused on how to apply this to the Re(z) boundary if this method is even correct.










    share|cite|improve this question


























      up vote
      0
      down vote

      favorite









      up vote
      0
      down vote

      favorite











      I have no clue if I am doing this correctly, just basing it off what my teacher did for a similar example.



      I look at image of $y=a$



      $u(x,y) = e^x cos{a}$



      $v(x,y) = e^x sin{a}$



      $$y=-pi , f(z) = u+iv = -e^x <0$$
      $$y=pi , f(z) = u+iv = -e^x <0$$



      So this is below the x-axis, I am confused on how to apply this to the Re(z) boundary if this method is even correct.










      share|cite|improve this question















      I have no clue if I am doing this correctly, just basing it off what my teacher did for a similar example.



      I look at image of $y=a$



      $u(x,y) = e^x cos{a}$



      $v(x,y) = e^x sin{a}$



      $$y=-pi , f(z) = u+iv = -e^x <0$$
      $$y=pi , f(z) = u+iv = -e^x <0$$



      So this is below the x-axis, I am confused on how to apply this to the Re(z) boundary if this method is even correct.







      complex-analysis






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      edited Nov 16 at 12:58









      amWhy

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      191k27223438










      asked Nov 16 at 11:36









      jaiidong

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      104






















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          Let $z = a+ib in mathbb{C}$.
          Then $Re(z) = a$, $Im(z) = b$ and $0 leq a leq 1/2$, $-pi leq b leq pi$.
          We know that $e^z = e^{a+ib} = e^a(cos(b) + isin(b))$.
          So, if we fix $a = 0$, for example, we have $e^a = 1$ and we are left with $cos(b) + isin(b)$, where $-pi leq b leq pi$.
          So, if this is the range of $b$, what figure you obtain?



          Now note that for every other $a$ you fix, you will obtain the same figure, but with a small difference.
          What's this difference?






          share|cite|improve this answer





















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            Let $z = a+ib in mathbb{C}$.
            Then $Re(z) = a$, $Im(z) = b$ and $0 leq a leq 1/2$, $-pi leq b leq pi$.
            We know that $e^z = e^{a+ib} = e^a(cos(b) + isin(b))$.
            So, if we fix $a = 0$, for example, we have $e^a = 1$ and we are left with $cos(b) + isin(b)$, where $-pi leq b leq pi$.
            So, if this is the range of $b$, what figure you obtain?



            Now note that for every other $a$ you fix, you will obtain the same figure, but with a small difference.
            What's this difference?






            share|cite|improve this answer

























              up vote
              1
              down vote













              Let $z = a+ib in mathbb{C}$.
              Then $Re(z) = a$, $Im(z) = b$ and $0 leq a leq 1/2$, $-pi leq b leq pi$.
              We know that $e^z = e^{a+ib} = e^a(cos(b) + isin(b))$.
              So, if we fix $a = 0$, for example, we have $e^a = 1$ and we are left with $cos(b) + isin(b)$, where $-pi leq b leq pi$.
              So, if this is the range of $b$, what figure you obtain?



              Now note that for every other $a$ you fix, you will obtain the same figure, but with a small difference.
              What's this difference?






              share|cite|improve this answer























                up vote
                1
                down vote










                up vote
                1
                down vote









                Let $z = a+ib in mathbb{C}$.
                Then $Re(z) = a$, $Im(z) = b$ and $0 leq a leq 1/2$, $-pi leq b leq pi$.
                We know that $e^z = e^{a+ib} = e^a(cos(b) + isin(b))$.
                So, if we fix $a = 0$, for example, we have $e^a = 1$ and we are left with $cos(b) + isin(b)$, where $-pi leq b leq pi$.
                So, if this is the range of $b$, what figure you obtain?



                Now note that for every other $a$ you fix, you will obtain the same figure, but with a small difference.
                What's this difference?






                share|cite|improve this answer












                Let $z = a+ib in mathbb{C}$.
                Then $Re(z) = a$, $Im(z) = b$ and $0 leq a leq 1/2$, $-pi leq b leq pi$.
                We know that $e^z = e^{a+ib} = e^a(cos(b) + isin(b))$.
                So, if we fix $a = 0$, for example, we have $e^a = 1$ and we are left with $cos(b) + isin(b)$, where $-pi leq b leq pi$.
                So, if this is the range of $b$, what figure you obtain?



                Now note that for every other $a$ you fix, you will obtain the same figure, but with a small difference.
                What's this difference?







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Nov 16 at 12:35









                Bias of Priene

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