Convert $(1+i) ^ {1+i}$ to polar form












1














Can someone please help me understand the exponent/logarithm relationships to get through this problem?
Thank you.






algebra-precalculus complex-numbers exponential-function







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  • 2




    $$z^wequiv e^{wlog(z)}implies (1+i)^{1+i}=e^{(1+i)log(1+i)}$$and$$log(z)=text{Log}(|z|)+iarg(z)$$where $arg(z)$ is the multivalued argument of $z$.
    – Mark Viola
    Dec 8 at 23:00










  • @MarkViola I’m sorry but could you expand? Still not sure how to go about solving, thank you.
    – Sam
    Dec 8 at 23:05










  • math.stackexchange.com/questions/2298090/… and math.stackexchange.com/questions/189703/…
    – lab bhattacharjee
    Dec 10 at 9:01


















1














Can someone please help me understand the exponent/logarithm relationships to get through this problem?
Thank you.






algebra-precalculus complex-numbers exponential-function







share|cite|improve this question




















  • 2




    $$z^wequiv e^{wlog(z)}implies (1+i)^{1+i}=e^{(1+i)log(1+i)}$$and$$log(z)=text{Log}(|z|)+iarg(z)$$where $arg(z)$ is the multivalued argument of $z$.
    – Mark Viola
    Dec 8 at 23:00










  • @MarkViola I’m sorry but could you expand? Still not sure how to go about solving, thank you.
    – Sam
    Dec 8 at 23:05










  • math.stackexchange.com/questions/2298090/… and math.stackexchange.com/questions/189703/…
    – lab bhattacharjee
    Dec 10 at 9:01
















1












1








1


1





Can someone please help me understand the exponent/logarithm relationships to get through this problem?
Thank you.






algebra-precalculus complex-numbers exponential-function







share|cite|improve this question















Can someone please help me understand the exponent/logarithm relationships to get through this problem?
Thank you.






algebra-precalculus complex-numbers exponential-function




algebra-precalculus complex-numbers exponential-function






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




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edited Dec 8 at 23:32









Key Flex

7,46941232




7,46941232










asked Dec 8 at 22:57









Sam

111




111








  • 2




    $$z^wequiv e^{wlog(z)}implies (1+i)^{1+i}=e^{(1+i)log(1+i)}$$and$$log(z)=text{Log}(|z|)+iarg(z)$$where $arg(z)$ is the multivalued argument of $z$.
    – Mark Viola
    Dec 8 at 23:00










  • @MarkViola I’m sorry but could you expand? Still not sure how to go about solving, thank you.
    – Sam
    Dec 8 at 23:05










  • math.stackexchange.com/questions/2298090/… and math.stackexchange.com/questions/189703/…
    – lab bhattacharjee
    Dec 10 at 9:01
















  • 2




    $$z^wequiv e^{wlog(z)}implies (1+i)^{1+i}=e^{(1+i)log(1+i)}$$and$$log(z)=text{Log}(|z|)+iarg(z)$$where $arg(z)$ is the multivalued argument of $z$.
    – Mark Viola
    Dec 8 at 23:00










  • @MarkViola I’m sorry but could you expand? Still not sure how to go about solving, thank you.
    – Sam
    Dec 8 at 23:05










  • math.stackexchange.com/questions/2298090/… and math.stackexchange.com/questions/189703/…
    – lab bhattacharjee
    Dec 10 at 9:01










2




2




$$z^wequiv e^{wlog(z)}implies (1+i)^{1+i}=e^{(1+i)log(1+i)}$$and$$log(z)=text{Log}(|z|)+iarg(z)$$where $arg(z)$ is the multivalued argument of $z$.
– Mark Viola
Dec 8 at 23:00




$$z^wequiv e^{wlog(z)}implies (1+i)^{1+i}=e^{(1+i)log(1+i)}$$and$$log(z)=text{Log}(|z|)+iarg(z)$$where $arg(z)$ is the multivalued argument of $z$.
– Mark Viola
Dec 8 at 23:00












@MarkViola I’m sorry but could you expand? Still not sure how to go about solving, thank you.
– Sam
Dec 8 at 23:05




@MarkViola I’m sorry but could you expand? Still not sure how to go about solving, thank you.
– Sam
Dec 8 at 23:05












math.stackexchange.com/questions/2298090/… and math.stackexchange.com/questions/189703/…
– lab bhattacharjee
Dec 10 at 9:01






math.stackexchange.com/questions/2298090/… and math.stackexchange.com/questions/189703/…
– lab bhattacharjee
Dec 10 at 9:01












2 Answers
2






active

oldest

votes


















0














Hints to explore




  • Can you write $1+i$ in polar form $re^{itheta}$ with $r$ being a positive real?


  • Can you write $1+i$ in the form $e^{s+i theta}$ with $s=log(r)$ being a real number?


  • Then consider $left(e^{s+i theta}right)^{1+i} = e^{(s+itheta)(1+i)} = e^{(s-theta)+i(s+ theta)}$


  • Then remember that $e^{x+iy} = e^x cos(y) + i e^xsin(y)$



That will get you one value. But complex powers are not so conveniently defined and $re^{itheta} = re^{i(theta+2npi)}$ for all integers $n$. So you should consider what different values of $n$ do to your earlier answer






share|cite|improve this answer





























    0














    We will need the logarithm function. In the complex plane this involves a choice, and the argument is multi-valued. We choose the principal one, which is the one that we usually expect. For $z=re^{it}$, we have $log z=log |z|+it$.



    We need to write $1+i$ in polar form: $1+i=sqrt2,e^{ipi/4}$.



    By definition, begin{align}(1+i)^{1+i}&=exp((1+i)log(1+i))=exp((1+i)(logsqrt2+itfracpi4)\ \
    &=exp(tfrac12,(1+i)(log2+itfracpi2)=exp(tfrac12,(-tfracpi2+log 2 + i(tfracpi2+log 2))\ \
    &exp(-tfracpi4+tfrac12log 2 + i(tfracpi4+tfrac12log 2))\ \
    &=e^{-pi/4}sqrt2,expleft[ileft(tfracpi4+log sqrt2right)right].
    end{align}






    share|cite|improve this answer























    • Yes, I had first written the last line with cos and sin. Thanks for noticing.
      – Martin Argerami
      Dec 8 at 23:30











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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    0














    Hints to explore




    • Can you write $1+i$ in polar form $re^{itheta}$ with $r$ being a positive real?


    • Can you write $1+i$ in the form $e^{s+i theta}$ with $s=log(r)$ being a real number?


    • Then consider $left(e^{s+i theta}right)^{1+i} = e^{(s+itheta)(1+i)} = e^{(s-theta)+i(s+ theta)}$


    • Then remember that $e^{x+iy} = e^x cos(y) + i e^xsin(y)$



    That will get you one value. But complex powers are not so conveniently defined and $re^{itheta} = re^{i(theta+2npi)}$ for all integers $n$. So you should consider what different values of $n$ do to your earlier answer






    share|cite|improve this answer


























      0














      Hints to explore




      • Can you write $1+i$ in polar form $re^{itheta}$ with $r$ being a positive real?


      • Can you write $1+i$ in the form $e^{s+i theta}$ with $s=log(r)$ being a real number?


      • Then consider $left(e^{s+i theta}right)^{1+i} = e^{(s+itheta)(1+i)} = e^{(s-theta)+i(s+ theta)}$


      • Then remember that $e^{x+iy} = e^x cos(y) + i e^xsin(y)$



      That will get you one value. But complex powers are not so conveniently defined and $re^{itheta} = re^{i(theta+2npi)}$ for all integers $n$. So you should consider what different values of $n$ do to your earlier answer






      share|cite|improve this answer
























        0












        0








        0






        Hints to explore




        • Can you write $1+i$ in polar form $re^{itheta}$ with $r$ being a positive real?


        • Can you write $1+i$ in the form $e^{s+i theta}$ with $s=log(r)$ being a real number?


        • Then consider $left(e^{s+i theta}right)^{1+i} = e^{(s+itheta)(1+i)} = e^{(s-theta)+i(s+ theta)}$


        • Then remember that $e^{x+iy} = e^x cos(y) + i e^xsin(y)$



        That will get you one value. But complex powers are not so conveniently defined and $re^{itheta} = re^{i(theta+2npi)}$ for all integers $n$. So you should consider what different values of $n$ do to your earlier answer






        share|cite|improve this answer












        Hints to explore




        • Can you write $1+i$ in polar form $re^{itheta}$ with $r$ being a positive real?


        • Can you write $1+i$ in the form $e^{s+i theta}$ with $s=log(r)$ being a real number?


        • Then consider $left(e^{s+i theta}right)^{1+i} = e^{(s+itheta)(1+i)} = e^{(s-theta)+i(s+ theta)}$


        • Then remember that $e^{x+iy} = e^x cos(y) + i e^xsin(y)$



        That will get you one value. But complex powers are not so conveniently defined and $re^{itheta} = re^{i(theta+2npi)}$ for all integers $n$. So you should consider what different values of $n$ do to your earlier answer







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 8 at 23:27









        Henry

        98k475160




        98k475160























            0














            We will need the logarithm function. In the complex plane this involves a choice, and the argument is multi-valued. We choose the principal one, which is the one that we usually expect. For $z=re^{it}$, we have $log z=log |z|+it$.



            We need to write $1+i$ in polar form: $1+i=sqrt2,e^{ipi/4}$.



            By definition, begin{align}(1+i)^{1+i}&=exp((1+i)log(1+i))=exp((1+i)(logsqrt2+itfracpi4)\ \
            &=exp(tfrac12,(1+i)(log2+itfracpi2)=exp(tfrac12,(-tfracpi2+log 2 + i(tfracpi2+log 2))\ \
            &exp(-tfracpi4+tfrac12log 2 + i(tfracpi4+tfrac12log 2))\ \
            &=e^{-pi/4}sqrt2,expleft[ileft(tfracpi4+log sqrt2right)right].
            end{align}






            share|cite|improve this answer























            • Yes, I had first written the last line with cos and sin. Thanks for noticing.
              – Martin Argerami
              Dec 8 at 23:30
















            0














            We will need the logarithm function. In the complex plane this involves a choice, and the argument is multi-valued. We choose the principal one, which is the one that we usually expect. For $z=re^{it}$, we have $log z=log |z|+it$.



            We need to write $1+i$ in polar form: $1+i=sqrt2,e^{ipi/4}$.



            By definition, begin{align}(1+i)^{1+i}&=exp((1+i)log(1+i))=exp((1+i)(logsqrt2+itfracpi4)\ \
            &=exp(tfrac12,(1+i)(log2+itfracpi2)=exp(tfrac12,(-tfracpi2+log 2 + i(tfracpi2+log 2))\ \
            &exp(-tfracpi4+tfrac12log 2 + i(tfracpi4+tfrac12log 2))\ \
            &=e^{-pi/4}sqrt2,expleft[ileft(tfracpi4+log sqrt2right)right].
            end{align}






            share|cite|improve this answer























            • Yes, I had first written the last line with cos and sin. Thanks for noticing.
              – Martin Argerami
              Dec 8 at 23:30














            0












            0








            0






            We will need the logarithm function. In the complex plane this involves a choice, and the argument is multi-valued. We choose the principal one, which is the one that we usually expect. For $z=re^{it}$, we have $log z=log |z|+it$.



            We need to write $1+i$ in polar form: $1+i=sqrt2,e^{ipi/4}$.



            By definition, begin{align}(1+i)^{1+i}&=exp((1+i)log(1+i))=exp((1+i)(logsqrt2+itfracpi4)\ \
            &=exp(tfrac12,(1+i)(log2+itfracpi2)=exp(tfrac12,(-tfracpi2+log 2 + i(tfracpi2+log 2))\ \
            &exp(-tfracpi4+tfrac12log 2 + i(tfracpi4+tfrac12log 2))\ \
            &=e^{-pi/4}sqrt2,expleft[ileft(tfracpi4+log sqrt2right)right].
            end{align}






            share|cite|improve this answer














            We will need the logarithm function. In the complex plane this involves a choice, and the argument is multi-valued. We choose the principal one, which is the one that we usually expect. For $z=re^{it}$, we have $log z=log |z|+it$.



            We need to write $1+i$ in polar form: $1+i=sqrt2,e^{ipi/4}$.



            By definition, begin{align}(1+i)^{1+i}&=exp((1+i)log(1+i))=exp((1+i)(logsqrt2+itfracpi4)\ \
            &=exp(tfrac12,(1+i)(log2+itfracpi2)=exp(tfrac12,(-tfracpi2+log 2 + i(tfracpi2+log 2))\ \
            &exp(-tfracpi4+tfrac12log 2 + i(tfracpi4+tfrac12log 2))\ \
            &=e^{-pi/4}sqrt2,expleft[ileft(tfracpi4+log sqrt2right)right].
            end{align}







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Dec 8 at 23:29

























            answered Dec 8 at 23:26









            Martin Argerami

            124k1176174




            124k1176174












            • Yes, I had first written the last line with cos and sin. Thanks for noticing.
              – Martin Argerami
              Dec 8 at 23:30


















            • Yes, I had first written the last line with cos and sin. Thanks for noticing.
              – Martin Argerami
              Dec 8 at 23:30
















            Yes, I had first written the last line with cos and sin. Thanks for noticing.
            – Martin Argerami
            Dec 8 at 23:30




            Yes, I had first written the last line with cos and sin. Thanks for noticing.
            – Martin Argerami
            Dec 8 at 23:30


















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