How can I convert $(a + bi)^{c+di}$ to polar form?












1












$begingroup$


Specifically, I need to convert $(-8i)^{2+πi/3}$ to polar form.
I understand I need to use Euler's formula, $e^{theta i} =cos theta + i sin theta$, but I'm not sure about the full process.



Thanks.






complex-numbers exponentiation polar-coordinates







share|cite|improve this question











$endgroup$












  • $begingroup$
    wolframalpha.com/input/?i=(-8+I)%5E(2%2BPi%2F3)
    $endgroup$
    – ablmf
    Dec 20 '18 at 0:49






  • 1




    $begingroup$
    The function $z^w$ in defined as $z^w=e^{wlog(z)}$ and is multivalued since $log(z)=text{Log}(|z|)+iarg(z)$ where $arg(z)$is the multivalued argument of $z$.
    $endgroup$
    – Mark Viola
    Dec 20 '18 at 0:52












  • $begingroup$
    I am not sure what that means.
    $endgroup$
    – Sam
    Dec 20 '18 at 0:55










  • $begingroup$
    Have a look at this post and upvote if you like it. I detailed $(4+5i)^{2-3i}$ and explained about principal value and multiplicative factors. math.stackexchange.com/questions/2446611/…
    $endgroup$
    – zwim
    Dec 20 '18 at 0:58
















1












$begingroup$


Specifically, I need to convert $(-8i)^{2+πi/3}$ to polar form.
I understand I need to use Euler's formula, $e^{theta i} =cos theta + i sin theta$, but I'm not sure about the full process.



Thanks.






complex-numbers exponentiation polar-coordinates







share|cite|improve this question











$endgroup$












  • $begingroup$
    wolframalpha.com/input/?i=(-8+I)%5E(2%2BPi%2F3)
    $endgroup$
    – ablmf
    Dec 20 '18 at 0:49






  • 1




    $begingroup$
    The function $z^w$ in defined as $z^w=e^{wlog(z)}$ and is multivalued since $log(z)=text{Log}(|z|)+iarg(z)$ where $arg(z)$is the multivalued argument of $z$.
    $endgroup$
    – Mark Viola
    Dec 20 '18 at 0:52












  • $begingroup$
    I am not sure what that means.
    $endgroup$
    – Sam
    Dec 20 '18 at 0:55










  • $begingroup$
    Have a look at this post and upvote if you like it. I detailed $(4+5i)^{2-3i}$ and explained about principal value and multiplicative factors. math.stackexchange.com/questions/2446611/…
    $endgroup$
    – zwim
    Dec 20 '18 at 0:58














1












1








1





$begingroup$


Specifically, I need to convert $(-8i)^{2+πi/3}$ to polar form.
I understand I need to use Euler's formula, $e^{theta i} =cos theta + i sin theta$, but I'm not sure about the full process.



Thanks.






complex-numbers exponentiation polar-coordinates







share|cite|improve this question











$endgroup$




Specifically, I need to convert $(-8i)^{2+πi/3}$ to polar form.
I understand I need to use Euler's formula, $e^{theta i} =cos theta + i sin theta$, but I'm not sure about the full process.



Thanks.






complex-numbers exponentiation polar-coordinates




complex-numbers exponentiation polar-coordinates






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 20 '18 at 7:09









Eric Wofsey

184k13211338




184k13211338










asked Dec 20 '18 at 0:44









SamSam

111




111












  • $begingroup$
    wolframalpha.com/input/?i=(-8+I)%5E(2%2BPi%2F3)
    $endgroup$
    – ablmf
    Dec 20 '18 at 0:49






  • 1




    $begingroup$
    The function $z^w$ in defined as $z^w=e^{wlog(z)}$ and is multivalued since $log(z)=text{Log}(|z|)+iarg(z)$ where $arg(z)$is the multivalued argument of $z$.
    $endgroup$
    – Mark Viola
    Dec 20 '18 at 0:52












  • $begingroup$
    I am not sure what that means.
    $endgroup$
    – Sam
    Dec 20 '18 at 0:55










  • $begingroup$
    Have a look at this post and upvote if you like it. I detailed $(4+5i)^{2-3i}$ and explained about principal value and multiplicative factors. math.stackexchange.com/questions/2446611/…
    $endgroup$
    – zwim
    Dec 20 '18 at 0:58


















  • $begingroup$
    wolframalpha.com/input/?i=(-8+I)%5E(2%2BPi%2F3)
    $endgroup$
    – ablmf
    Dec 20 '18 at 0:49






  • 1




    $begingroup$
    The function $z^w$ in defined as $z^w=e^{wlog(z)}$ and is multivalued since $log(z)=text{Log}(|z|)+iarg(z)$ where $arg(z)$is the multivalued argument of $z$.
    $endgroup$
    – Mark Viola
    Dec 20 '18 at 0:52












  • $begingroup$
    I am not sure what that means.
    $endgroup$
    – Sam
    Dec 20 '18 at 0:55










  • $begingroup$
    Have a look at this post and upvote if you like it. I detailed $(4+5i)^{2-3i}$ and explained about principal value and multiplicative factors. math.stackexchange.com/questions/2446611/…
    $endgroup$
    – zwim
    Dec 20 '18 at 0:58
















$begingroup$
wolframalpha.com/input/?i=(-8+I)%5E(2%2BPi%2F3)
$endgroup$
– ablmf
Dec 20 '18 at 0:49




$begingroup$
wolframalpha.com/input/?i=(-8+I)%5E(2%2BPi%2F3)
$endgroup$
– ablmf
Dec 20 '18 at 0:49




1




1




$begingroup$
The function $z^w$ in defined as $z^w=e^{wlog(z)}$ and is multivalued since $log(z)=text{Log}(|z|)+iarg(z)$ where $arg(z)$is the multivalued argument of $z$.
$endgroup$
– Mark Viola
Dec 20 '18 at 0:52






$begingroup$
The function $z^w$ in defined as $z^w=e^{wlog(z)}$ and is multivalued since $log(z)=text{Log}(|z|)+iarg(z)$ where $arg(z)$is the multivalued argument of $z$.
$endgroup$
– Mark Viola
Dec 20 '18 at 0:52














$begingroup$
I am not sure what that means.
$endgroup$
– Sam
Dec 20 '18 at 0:55




$begingroup$
I am not sure what that means.
$endgroup$
– Sam
Dec 20 '18 at 0:55












$begingroup$
Have a look at this post and upvote if you like it. I detailed $(4+5i)^{2-3i}$ and explained about principal value and multiplicative factors. math.stackexchange.com/questions/2446611/…
$endgroup$
– zwim
Dec 20 '18 at 0:58




$begingroup$
Have a look at this post and upvote if you like it. I detailed $(4+5i)^{2-3i}$ and explained about principal value and multiplicative factors. math.stackexchange.com/questions/2446611/…
$endgroup$
– zwim
Dec 20 '18 at 0:58










1 Answer
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1












$begingroup$

For any $z,w in mathbb{C}$, with $z = a + bi$ and $w = c+di$, with $a,b,c,d in mathbb{R}$.



Then:



$$begin{split}
z^w &= mathrm{e}^{log z^w} \
&= mathrm{e}^{w log z} \
&= mathrm{e}^{w (log |z| + iarg z)} \
&= mathrm{e}^{(c + di) (log |z| + iarg z)} \
&= mathrm{e}^{(c log |z| - d arg z+ i(carg z + d log |z|))} \
&= mathrm{e}^{(c log |z| - d arg z)} mathrm{e}^{i(carg z + d log |z|)} \
end{split}
$$



In this last line, the magnitude is $mathrm{e}^{(c log |z| - d arg z)}$ and the angle is $(carg z + d log |z|)$



For your specific case: $z = -8i$ and $w = 2 + pi/3 i$, so $|z| = 8$, $arg z = -pi/2 + 2pi k , (text{with} , k in mathbb{Z})$, $c = 2$, and $d = pi/3$.



So (taking $k = 0$ for a single example):



$$begin{split}
z^w &= mathrm{e}^{(c log |z| - d arg z)} mathrm{e}^{i(carg z + d log |z|)} \
&= mathrm{e}^{(2 times log 8 - pi/3 times -pi/2)} mathrm{e}^{i(2 times -pi/2 + pi/3 times log 8)} \
&= mathrm{e}^{(2 log 8 + pi^2/6)} mathrm{e}^{i(-pi+ pi/3 log 8)} \
&approx 331.56277234 mathrm{e}^{-0.964006563 i} \
end{split}$$



This is only the principal log though, there are an infinite number of possible solutions.






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    1 Answer
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    1 Answer
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    active

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    1












    $begingroup$

    For any $z,w in mathbb{C}$, with $z = a + bi$ and $w = c+di$, with $a,b,c,d in mathbb{R}$.



    Then:



    $$begin{split}
    z^w &= mathrm{e}^{log z^w} \
    &= mathrm{e}^{w log z} \
    &= mathrm{e}^{w (log |z| + iarg z)} \
    &= mathrm{e}^{(c + di) (log |z| + iarg z)} \
    &= mathrm{e}^{(c log |z| - d arg z+ i(carg z + d log |z|))} \
    &= mathrm{e}^{(c log |z| - d arg z)} mathrm{e}^{i(carg z + d log |z|)} \
    end{split}
    $$



    In this last line, the magnitude is $mathrm{e}^{(c log |z| - d arg z)}$ and the angle is $(carg z + d log |z|)$



    For your specific case: $z = -8i$ and $w = 2 + pi/3 i$, so $|z| = 8$, $arg z = -pi/2 + 2pi k , (text{with} , k in mathbb{Z})$, $c = 2$, and $d = pi/3$.



    So (taking $k = 0$ for a single example):



    $$begin{split}
    z^w &= mathrm{e}^{(c log |z| - d arg z)} mathrm{e}^{i(carg z + d log |z|)} \
    &= mathrm{e}^{(2 times log 8 - pi/3 times -pi/2)} mathrm{e}^{i(2 times -pi/2 + pi/3 times log 8)} \
    &= mathrm{e}^{(2 log 8 + pi^2/6)} mathrm{e}^{i(-pi+ pi/3 log 8)} \
    &approx 331.56277234 mathrm{e}^{-0.964006563 i} \
    end{split}$$



    This is only the principal log though, there are an infinite number of possible solutions.






    share|cite|improve this answer











    $endgroup$


















      1












      $begingroup$

      For any $z,w in mathbb{C}$, with $z = a + bi$ and $w = c+di$, with $a,b,c,d in mathbb{R}$.



      Then:



      $$begin{split}
      z^w &= mathrm{e}^{log z^w} \
      &= mathrm{e}^{w log z} \
      &= mathrm{e}^{w (log |z| + iarg z)} \
      &= mathrm{e}^{(c + di) (log |z| + iarg z)} \
      &= mathrm{e}^{(c log |z| - d arg z+ i(carg z + d log |z|))} \
      &= mathrm{e}^{(c log |z| - d arg z)} mathrm{e}^{i(carg z + d log |z|)} \
      end{split}
      $$



      In this last line, the magnitude is $mathrm{e}^{(c log |z| - d arg z)}$ and the angle is $(carg z + d log |z|)$



      For your specific case: $z = -8i$ and $w = 2 + pi/3 i$, so $|z| = 8$, $arg z = -pi/2 + 2pi k , (text{with} , k in mathbb{Z})$, $c = 2$, and $d = pi/3$.



      So (taking $k = 0$ for a single example):



      $$begin{split}
      z^w &= mathrm{e}^{(c log |z| - d arg z)} mathrm{e}^{i(carg z + d log |z|)} \
      &= mathrm{e}^{(2 times log 8 - pi/3 times -pi/2)} mathrm{e}^{i(2 times -pi/2 + pi/3 times log 8)} \
      &= mathrm{e}^{(2 log 8 + pi^2/6)} mathrm{e}^{i(-pi+ pi/3 log 8)} \
      &approx 331.56277234 mathrm{e}^{-0.964006563 i} \
      end{split}$$



      This is only the principal log though, there are an infinite number of possible solutions.






      share|cite|improve this answer











      $endgroup$
















        1












        1








        1





        $begingroup$

        For any $z,w in mathbb{C}$, with $z = a + bi$ and $w = c+di$, with $a,b,c,d in mathbb{R}$.



        Then:



        $$begin{split}
        z^w &= mathrm{e}^{log z^w} \
        &= mathrm{e}^{w log z} \
        &= mathrm{e}^{w (log |z| + iarg z)} \
        &= mathrm{e}^{(c + di) (log |z| + iarg z)} \
        &= mathrm{e}^{(c log |z| - d arg z+ i(carg z + d log |z|))} \
        &= mathrm{e}^{(c log |z| - d arg z)} mathrm{e}^{i(carg z + d log |z|)} \
        end{split}
        $$



        In this last line, the magnitude is $mathrm{e}^{(c log |z| - d arg z)}$ and the angle is $(carg z + d log |z|)$



        For your specific case: $z = -8i$ and $w = 2 + pi/3 i$, so $|z| = 8$, $arg z = -pi/2 + 2pi k , (text{with} , k in mathbb{Z})$, $c = 2$, and $d = pi/3$.



        So (taking $k = 0$ for a single example):



        $$begin{split}
        z^w &= mathrm{e}^{(c log |z| - d arg z)} mathrm{e}^{i(carg z + d log |z|)} \
        &= mathrm{e}^{(2 times log 8 - pi/3 times -pi/2)} mathrm{e}^{i(2 times -pi/2 + pi/3 times log 8)} \
        &= mathrm{e}^{(2 log 8 + pi^2/6)} mathrm{e}^{i(-pi+ pi/3 log 8)} \
        &approx 331.56277234 mathrm{e}^{-0.964006563 i} \
        end{split}$$



        This is only the principal log though, there are an infinite number of possible solutions.






        share|cite|improve this answer











        $endgroup$



        For any $z,w in mathbb{C}$, with $z = a + bi$ and $w = c+di$, with $a,b,c,d in mathbb{R}$.



        Then:



        $$begin{split}
        z^w &= mathrm{e}^{log z^w} \
        &= mathrm{e}^{w log z} \
        &= mathrm{e}^{w (log |z| + iarg z)} \
        &= mathrm{e}^{(c + di) (log |z| + iarg z)} \
        &= mathrm{e}^{(c log |z| - d arg z+ i(carg z + d log |z|))} \
        &= mathrm{e}^{(c log |z| - d arg z)} mathrm{e}^{i(carg z + d log |z|)} \
        end{split}
        $$



        In this last line, the magnitude is $mathrm{e}^{(c log |z| - d arg z)}$ and the angle is $(carg z + d log |z|)$



        For your specific case: $z = -8i$ and $w = 2 + pi/3 i$, so $|z| = 8$, $arg z = -pi/2 + 2pi k , (text{with} , k in mathbb{Z})$, $c = 2$, and $d = pi/3$.



        So (taking $k = 0$ for a single example):



        $$begin{split}
        z^w &= mathrm{e}^{(c log |z| - d arg z)} mathrm{e}^{i(carg z + d log |z|)} \
        &= mathrm{e}^{(2 times log 8 - pi/3 times -pi/2)} mathrm{e}^{i(2 times -pi/2 + pi/3 times log 8)} \
        &= mathrm{e}^{(2 log 8 + pi^2/6)} mathrm{e}^{i(-pi+ pi/3 log 8)} \
        &approx 331.56277234 mathrm{e}^{-0.964006563 i} \
        end{split}$$



        This is only the principal log though, there are an infinite number of possible solutions.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 20 '18 at 5:18

























        answered Dec 20 '18 at 1:24









        OmnipotentEntityOmnipotentEntity

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        462318






























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