Solving $z^{1+i}=4$.












1














Solution:



Let $z=re^{itheta}=r(cos theta + isin theta)$. Then $z^{1+i}=e^{itheta(1+i)}=e^{itheta -theta}=4$. So $e^{itheta-theta}=e^{-theta}e^{itheta}=e^{-theta}(cos theta + isin theta)=4(cos 0 +isin 0) Longleftrightarrow e^{-theta}=4 quad text{and}quad theta =0$. But, $e^0 =1$. What am I doing wrong?



EDIT 10/26/18
Would like to provide a more systematic and traditional approach:
begin{align}
r^{1+i}e^{itheta(1+i)}&=r^{1+i}e^{-theta}(costheta + isintheta)=4(cos 0 +isin 0)=4\
r^{1+i}e^{-theta}&=4 quad text{and}quad theta=0\
log_e(r^{1+i}e^{-theta})&=log_{e}(4)\
log_(r^{1+i})+log(e^{-theta})&=log_{e}(4)\
(1+i)log_{e}(r)-theta&=log_{e}(4)\
log_{e}(r)&=frac{log_e(4)}{1+i}\
log_{e}(r)&=log_{e}(4^{frac{1}{1+i}})\
r&=4^{frac{1}{1+i}}\
r&=4^{frac{1-i}{2}}=[4^frac{1}{2}]^{1-i}=2^{1-i}
end{align}






complex-numbers







share|cite|improve this question




















  • 1




    You completely forgot the $r^{1+i}$ part...
    – b00n heT
    Oct 24 '18 at 6:11








  • 1




    $z^{1+i}=r^{(1+i)} e^{itheta(1+i)}$
    – G.H.lee
    Oct 24 '18 at 6:11






  • 1




    What happened to $r$?
    – Carsten S
    Oct 24 '18 at 6:11










  • thanks. I'll redo it.
    – TheLast Cipher
    Oct 24 '18 at 6:12
















1














Solution:



Let $z=re^{itheta}=r(cos theta + isin theta)$. Then $z^{1+i}=e^{itheta(1+i)}=e^{itheta -theta}=4$. So $e^{itheta-theta}=e^{-theta}e^{itheta}=e^{-theta}(cos theta + isin theta)=4(cos 0 +isin 0) Longleftrightarrow e^{-theta}=4 quad text{and}quad theta =0$. But, $e^0 =1$. What am I doing wrong?



EDIT 10/26/18
Would like to provide a more systematic and traditional approach:
begin{align}
r^{1+i}e^{itheta(1+i)}&=r^{1+i}e^{-theta}(costheta + isintheta)=4(cos 0 +isin 0)=4\
r^{1+i}e^{-theta}&=4 quad text{and}quad theta=0\
log_e(r^{1+i}e^{-theta})&=log_{e}(4)\
log_(r^{1+i})+log(e^{-theta})&=log_{e}(4)\
(1+i)log_{e}(r)-theta&=log_{e}(4)\
log_{e}(r)&=frac{log_e(4)}{1+i}\
log_{e}(r)&=log_{e}(4^{frac{1}{1+i}})\
r&=4^{frac{1}{1+i}}\
r&=4^{frac{1-i}{2}}=[4^frac{1}{2}]^{1-i}=2^{1-i}
end{align}






complex-numbers







share|cite|improve this question




















  • 1




    You completely forgot the $r^{1+i}$ part...
    – b00n heT
    Oct 24 '18 at 6:11








  • 1




    $z^{1+i}=r^{(1+i)} e^{itheta(1+i)}$
    – G.H.lee
    Oct 24 '18 at 6:11






  • 1




    What happened to $r$?
    – Carsten S
    Oct 24 '18 at 6:11










  • thanks. I'll redo it.
    – TheLast Cipher
    Oct 24 '18 at 6:12














1












1








1







Solution:



Let $z=re^{itheta}=r(cos theta + isin theta)$. Then $z^{1+i}=e^{itheta(1+i)}=e^{itheta -theta}=4$. So $e^{itheta-theta}=e^{-theta}e^{itheta}=e^{-theta}(cos theta + isin theta)=4(cos 0 +isin 0) Longleftrightarrow e^{-theta}=4 quad text{and}quad theta =0$. But, $e^0 =1$. What am I doing wrong?



EDIT 10/26/18
Would like to provide a more systematic and traditional approach:
begin{align}
r^{1+i}e^{itheta(1+i)}&=r^{1+i}e^{-theta}(costheta + isintheta)=4(cos 0 +isin 0)=4\
r^{1+i}e^{-theta}&=4 quad text{and}quad theta=0\
log_e(r^{1+i}e^{-theta})&=log_{e}(4)\
log_(r^{1+i})+log(e^{-theta})&=log_{e}(4)\
(1+i)log_{e}(r)-theta&=log_{e}(4)\
log_{e}(r)&=frac{log_e(4)}{1+i}\
log_{e}(r)&=log_{e}(4^{frac{1}{1+i}})\
r&=4^{frac{1}{1+i}}\
r&=4^{frac{1-i}{2}}=[4^frac{1}{2}]^{1-i}=2^{1-i}
end{align}






complex-numbers







share|cite|improve this question















Solution:



Let $z=re^{itheta}=r(cos theta + isin theta)$. Then $z^{1+i}=e^{itheta(1+i)}=e^{itheta -theta}=4$. So $e^{itheta-theta}=e^{-theta}e^{itheta}=e^{-theta}(cos theta + isin theta)=4(cos 0 +isin 0) Longleftrightarrow e^{-theta}=4 quad text{and}quad theta =0$. But, $e^0 =1$. What am I doing wrong?



EDIT 10/26/18
Would like to provide a more systematic and traditional approach:
begin{align}
r^{1+i}e^{itheta(1+i)}&=r^{1+i}e^{-theta}(costheta + isintheta)=4(cos 0 +isin 0)=4\
r^{1+i}e^{-theta}&=4 quad text{and}quad theta=0\
log_e(r^{1+i}e^{-theta})&=log_{e}(4)\
log_(r^{1+i})+log(e^{-theta})&=log_{e}(4)\
(1+i)log_{e}(r)-theta&=log_{e}(4)\
log_{e}(r)&=frac{log_e(4)}{1+i}\
log_{e}(r)&=log_{e}(4^{frac{1}{1+i}})\
r&=4^{frac{1}{1+i}}\
r&=4^{frac{1-i}{2}}=[4^frac{1}{2}]^{1-i}=2^{1-i}
end{align}






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 Oct 26 '18 at 5:16

























asked Oct 24 '18 at 6:08









TheLast Cipher

690715




690715








  • 1




    You completely forgot the $r^{1+i}$ part...
    – b00n heT
    Oct 24 '18 at 6:11








  • 1




    $z^{1+i}=r^{(1+i)} e^{itheta(1+i)}$
    – G.H.lee
    Oct 24 '18 at 6:11






  • 1




    What happened to $r$?
    – Carsten S
    Oct 24 '18 at 6:11










  • thanks. I'll redo it.
    – TheLast Cipher
    Oct 24 '18 at 6:12














  • 1




    You completely forgot the $r^{1+i}$ part...
    – b00n heT
    Oct 24 '18 at 6:11








  • 1




    $z^{1+i}=r^{(1+i)} e^{itheta(1+i)}$
    – G.H.lee
    Oct 24 '18 at 6:11






  • 1




    What happened to $r$?
    – Carsten S
    Oct 24 '18 at 6:11










  • thanks. I'll redo it.
    – TheLast Cipher
    Oct 24 '18 at 6:12








1




1




You completely forgot the $r^{1+i}$ part...
– b00n heT
Oct 24 '18 at 6:11






You completely forgot the $r^{1+i}$ part...
– b00n heT
Oct 24 '18 at 6:11






1




1




$z^{1+i}=r^{(1+i)} e^{itheta(1+i)}$
– G.H.lee
Oct 24 '18 at 6:11




$z^{1+i}=r^{(1+i)} e^{itheta(1+i)}$
– G.H.lee
Oct 24 '18 at 6:11




1




1




What happened to $r$?
– Carsten S
Oct 24 '18 at 6:11




What happened to $r$?
– Carsten S
Oct 24 '18 at 6:11












thanks. I'll redo it.
– TheLast Cipher
Oct 24 '18 at 6:12




thanks. I'll redo it.
– TheLast Cipher
Oct 24 '18 at 6:12










3 Answers
3






active

oldest

votes


















1














$$4 = 2^2 = z^{1 + i}$$



$$2 log 2 = (1+i) log z$$



$${2 over 1+i} log 2 = log z$$



$$2^{2 over 1 + i} = z$$



$$2^{2(1-i) over (1+i)(1-i)}$$



$$2^{2(1-i) over 2} = z$$



$$2^{1 - i} = z$$



Check:



$$left( 2^{1-i}right)^{1+i} = 2^{1+1} = 2^2 = 4$$






share|cite|improve this answer























  • $$pm2=z^{(1+i)/2},(pm2)^{1-i}=?$$
    – lab bhattacharjee
    Oct 24 '18 at 6:37



















2














$4=2^2$



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



$z^{1+i}=2^{(1-i)(1+i)}$



Taking $1+i$ th root of both sides we get:



$z=2^{1-i}$






share|cite|improve this answer





























    0














    A systematic way would go like this. Let the power function be defined as $z^a = e^{a ln z}$ and the logarithm be defined as $ln z = ln |z| + i arg z, , -pi < arg z leq pi$. $e^z$ is periodic with period $2 pi i$, therefore
    $$z^c = w ,Leftrightarrow,
    e^{c ln z} = e^{ln w} ,Leftrightarrow,
    c ln z = ln w + 2 pi i k ,Leftrightarrow \
    ln |z| + i arg z =
    operatorname{Re} zeta + i operatorname{Im} zeta ,Leftrightarrow \
    z = e^zeta land -pi < operatorname{Im} zeta leq pi, \
    text{where } zeta = frac {ln w + 2 pi i k} c,
    ;k in mathbb Z.$$
    The number of solutions will depend on $c$ and $w$. Substituting $c = 1 + i, , w = 4$ gives $-pi < pi k - ln 2 leq pi$, therefore there are two solutions $z = 2^{1 - i}$ and $z = -2^{1 - i} e^pi$, corresponding to $k = 0$ and $k = 1$.






    share|cite|improve this answer





















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      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      1














      $$4 = 2^2 = z^{1 + i}$$



      $$2 log 2 = (1+i) log z$$



      $${2 over 1+i} log 2 = log z$$



      $$2^{2 over 1 + i} = z$$



      $$2^{2(1-i) over (1+i)(1-i)}$$



      $$2^{2(1-i) over 2} = z$$



      $$2^{1 - i} = z$$



      Check:



      $$left( 2^{1-i}right)^{1+i} = 2^{1+1} = 2^2 = 4$$






      share|cite|improve this answer























      • $$pm2=z^{(1+i)/2},(pm2)^{1-i}=?$$
        – lab bhattacharjee
        Oct 24 '18 at 6:37
















      1














      $$4 = 2^2 = z^{1 + i}$$



      $$2 log 2 = (1+i) log z$$



      $${2 over 1+i} log 2 = log z$$



      $$2^{2 over 1 + i} = z$$



      $$2^{2(1-i) over (1+i)(1-i)}$$



      $$2^{2(1-i) over 2} = z$$



      $$2^{1 - i} = z$$



      Check:



      $$left( 2^{1-i}right)^{1+i} = 2^{1+1} = 2^2 = 4$$






      share|cite|improve this answer























      • $$pm2=z^{(1+i)/2},(pm2)^{1-i}=?$$
        – lab bhattacharjee
        Oct 24 '18 at 6:37














      1












      1








      1






      $$4 = 2^2 = z^{1 + i}$$



      $$2 log 2 = (1+i) log z$$



      $${2 over 1+i} log 2 = log z$$



      $$2^{2 over 1 + i} = z$$



      $$2^{2(1-i) over (1+i)(1-i)}$$



      $$2^{2(1-i) over 2} = z$$



      $$2^{1 - i} = z$$



      Check:



      $$left( 2^{1-i}right)^{1+i} = 2^{1+1} = 2^2 = 4$$






      share|cite|improve this answer














      $$4 = 2^2 = z^{1 + i}$$



      $$2 log 2 = (1+i) log z$$



      $${2 over 1+i} log 2 = log z$$



      $$2^{2 over 1 + i} = z$$



      $$2^{2(1-i) over (1+i)(1-i)}$$



      $$2^{2(1-i) over 2} = z$$



      $$2^{1 - i} = z$$



      Check:



      $$left( 2^{1-i}right)^{1+i} = 2^{1+1} = 2^2 = 4$$







      share|cite|improve this answer














      share|cite|improve this answer



      share|cite|improve this answer








      edited Oct 24 '18 at 6:38

























      answered Oct 24 '18 at 6:31









      David G. Stork

      9,78721232




      9,78721232












      • $$pm2=z^{(1+i)/2},(pm2)^{1-i}=?$$
        – lab bhattacharjee
        Oct 24 '18 at 6:37


















      • $$pm2=z^{(1+i)/2},(pm2)^{1-i}=?$$
        – lab bhattacharjee
        Oct 24 '18 at 6:37
















      $$pm2=z^{(1+i)/2},(pm2)^{1-i}=?$$
      – lab bhattacharjee
      Oct 24 '18 at 6:37




      $$pm2=z^{(1+i)/2},(pm2)^{1-i}=?$$
      – lab bhattacharjee
      Oct 24 '18 at 6:37











      2














      $4=2^2$



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



      $z^{1+i}=2^{(1-i)(1+i)}$



      Taking $1+i$ th root of both sides we get:



      $z=2^{1-i}$






      share|cite|improve this answer


























        2














        $4=2^2$



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



        $z^{1+i}=2^{(1-i)(1+i)}$



        Taking $1+i$ th root of both sides we get:



        $z=2^{1-i}$






        share|cite|improve this answer
























          2












          2








          2






          $4=2^2$



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



          $z^{1+i}=2^{(1-i)(1+i)}$



          Taking $1+i$ th root of both sides we get:



          $z=2^{1-i}$






          share|cite|improve this answer












          $4=2^2$



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



          $z^{1+i}=2^{(1-i)(1+i)}$



          Taking $1+i$ th root of both sides we get:



          $z=2^{1-i}$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Oct 24 '18 at 6:39









          sirous

          1,5991513




          1,5991513























              0














              A systematic way would go like this. Let the power function be defined as $z^a = e^{a ln z}$ and the logarithm be defined as $ln z = ln |z| + i arg z, , -pi < arg z leq pi$. $e^z$ is periodic with period $2 pi i$, therefore
              $$z^c = w ,Leftrightarrow,
              e^{c ln z} = e^{ln w} ,Leftrightarrow,
              c ln z = ln w + 2 pi i k ,Leftrightarrow \
              ln |z| + i arg z =
              operatorname{Re} zeta + i operatorname{Im} zeta ,Leftrightarrow \
              z = e^zeta land -pi < operatorname{Im} zeta leq pi, \
              text{where } zeta = frac {ln w + 2 pi i k} c,
              ;k in mathbb Z.$$
              The number of solutions will depend on $c$ and $w$. Substituting $c = 1 + i, , w = 4$ gives $-pi < pi k - ln 2 leq pi$, therefore there are two solutions $z = 2^{1 - i}$ and $z = -2^{1 - i} e^pi$, corresponding to $k = 0$ and $k = 1$.






              share|cite|improve this answer


























                0














                A systematic way would go like this. Let the power function be defined as $z^a = e^{a ln z}$ and the logarithm be defined as $ln z = ln |z| + i arg z, , -pi < arg z leq pi$. $e^z$ is periodic with period $2 pi i$, therefore
                $$z^c = w ,Leftrightarrow,
                e^{c ln z} = e^{ln w} ,Leftrightarrow,
                c ln z = ln w + 2 pi i k ,Leftrightarrow \
                ln |z| + i arg z =
                operatorname{Re} zeta + i operatorname{Im} zeta ,Leftrightarrow \
                z = e^zeta land -pi < operatorname{Im} zeta leq pi, \
                text{where } zeta = frac {ln w + 2 pi i k} c,
                ;k in mathbb Z.$$
                The number of solutions will depend on $c$ and $w$. Substituting $c = 1 + i, , w = 4$ gives $-pi < pi k - ln 2 leq pi$, therefore there are two solutions $z = 2^{1 - i}$ and $z = -2^{1 - i} e^pi$, corresponding to $k = 0$ and $k = 1$.






                share|cite|improve this answer
























                  0












                  0








                  0






                  A systematic way would go like this. Let the power function be defined as $z^a = e^{a ln z}$ and the logarithm be defined as $ln z = ln |z| + i arg z, , -pi < arg z leq pi$. $e^z$ is periodic with period $2 pi i$, therefore
                  $$z^c = w ,Leftrightarrow,
                  e^{c ln z} = e^{ln w} ,Leftrightarrow,
                  c ln z = ln w + 2 pi i k ,Leftrightarrow \
                  ln |z| + i arg z =
                  operatorname{Re} zeta + i operatorname{Im} zeta ,Leftrightarrow \
                  z = e^zeta land -pi < operatorname{Im} zeta leq pi, \
                  text{where } zeta = frac {ln w + 2 pi i k} c,
                  ;k in mathbb Z.$$
                  The number of solutions will depend on $c$ and $w$. Substituting $c = 1 + i, , w = 4$ gives $-pi < pi k - ln 2 leq pi$, therefore there are two solutions $z = 2^{1 - i}$ and $z = -2^{1 - i} e^pi$, corresponding to $k = 0$ and $k = 1$.






                  share|cite|improve this answer












                  A systematic way would go like this. Let the power function be defined as $z^a = e^{a ln z}$ and the logarithm be defined as $ln z = ln |z| + i arg z, , -pi < arg z leq pi$. $e^z$ is periodic with period $2 pi i$, therefore
                  $$z^c = w ,Leftrightarrow,
                  e^{c ln z} = e^{ln w} ,Leftrightarrow,
                  c ln z = ln w + 2 pi i k ,Leftrightarrow \
                  ln |z| + i arg z =
                  operatorname{Re} zeta + i operatorname{Im} zeta ,Leftrightarrow \
                  z = e^zeta land -pi < operatorname{Im} zeta leq pi, \
                  text{where } zeta = frac {ln w + 2 pi i k} c,
                  ;k in mathbb Z.$$
                  The number of solutions will depend on $c$ and $w$. Substituting $c = 1 + i, , w = 4$ gives $-pi < pi k - ln 2 leq pi$, therefore there are two solutions $z = 2^{1 - i}$ and $z = -2^{1 - i} e^pi$, corresponding to $k = 0$ and $k = 1$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Nov 19 '18 at 4:15









                  Maxim

                  4,5031219




                  4,5031219






























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                      Can't locate Autom4te/ChannelDefs.pm in @INC (when it definitely is there)

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                      up vote 1 down vote favorite I've been running into trouble running make for a build process that I know works on a 32-bit Ubuntu VM. I am running a 64-bit Ubuntu VM, and I have a feeling that the 64-bit may be the problem, but am not entirely sure. Basically, when I run the make command, I get the following error: Can't locate Autom4te/ChannelDefs.pm in @INC (@INC contains: [...]/staging_dir/host/share/autoconf /etc/perl /usr/local/lib/perl/5.14.2 /usr/local/share/perl/5.14.2 /usr/lib/perl5 /usr/share/perl5 /usr/lib/perl/5.14 /usr/share/perl/5.14 /usr/local/lib/site_perl .) at [...]/staging_dir/host/bin/autoreconf line 40. Now if I navigate to [...]/staging_dir/host/share/autoconf I can see that, contrary to what autoreconf thinks, Autom4te/ChannelDefs.pm definitely exists, so I don't really understand what
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