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Module & argument of complex number exp(ix) +exp(2ix)

Show that $|z+z^2|=2 cos fractheta2$Finding argument to complex number?Solve cos(z) + sin(z) = i, where z is a complex number and i the imaginary unitFinding a Trigonometric Form of Complex NumberCosine double angle formula for complex number anglesTrigonometric presentation of a complex numberConvert to algebraic form the following complex numberHow to decompose a complex number into a sum of two unitary modulus complex numbers?Complex number proof involving trigonometryDetermine the polar form of $sintheta + i(1+ costheta)$

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2

1

$begingroup$

I tried unsuccessfully to solve the following complex expression and get the module and the argument.

$e^ix+e^2ix$

I converted the whole expression to trigonometric function cos and sin but it got more complex than it looks at the origin.

complex-numbers

share|cite|improve this question

asked Sep 20 at 8:14

SAM.AmSAM.Am

1851010 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Is the module the absolute value ? Your approach is good. Why do you expect an easy formula for module and argument ?
  $endgroup$
  – Peter
  Sep 20 at 8:24
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  for modulus, see this question
  $endgroup$
  – J. W. Tanner
  Sep 20 at 11:32
  

  
  
  
  

add a comment
 | 

2

1

$begingroup$

I tried unsuccessfully to solve the following complex expression and get the module and the argument.

$e^ix+e^2ix$

I converted the whole expression to trigonometric function cos and sin but it got more complex than it looks at the origin.

complex-numbers

share|cite|improve this question

asked Sep 20 at 8:14

SAM.AmSAM.Am

1851010 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Is the module the absolute value ? Your approach is good. Why do you expect an easy formula for module and argument ?
  $endgroup$
  – Peter
  Sep 20 at 8:24
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  for modulus, see this question
  $endgroup$
  – J. W. Tanner
  Sep 20 at 11:32
  

  
  
  
  

add a comment
 | 

$begingroup$

I tried unsuccessfully to solve the following complex expression and get the module and the argument.

$e^ix+e^2ix$

I converted the whole expression to trigonometric function cos and sin but it got more complex than it looks at the origin.

complex-numbers

share|cite|improve this question

asked Sep 20 at 8:14

SAM.AmSAM.Am

1851010 bronze badges

$endgroup$

share|cite|improve this question

asked Sep 20 at 8:14

SAM.AmSAM.Am

1851010 bronze badges

share|cite|improve this question

asked Sep 20 at 8:14

SAM.AmSAM.Am

1851010 bronze badges

asked Sep 20 at 8:14

SAM.AmSAM.Am

1851010 bronze badges

asked Sep 20 at 8:14

SAM.AmSAM.Am

1851010 bronze badges

3 Answers
3

active

oldest

votes

5

$begingroup$

$$
beginalign
e^ix+e^2ix
&=e^frac32ixleft(e^frac12ix+e^-frac12ixright)\
&=e^icolor#090frac32xcolor#C002cosleft(tfrac12xright)
endalign
$$
So
$$
left|,e^ix+e^2ix,right|=left|,color#C002cosleft(tfrac12xright),right|
$$
and
$$
argleft(e^ix+e^2ixright)equivcolor#090tfrac32x+pileft[,cosleft(tfrac12xright)lt0,right]pmod2pi
$$
where $[cdots]$ are Iverson Brackets.

share|cite|improve this answer

edited Sep 20 at 9:14

answered Sep 20 at 8:36

robjohn♦robjohn

283k2929 gold badges335335 silver badges667667 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  But $$cosdfrac x2$$ can be $$<0?$$
  $endgroup$
  – lab bhattacharjee
  Sep 20 at 8:37
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Very nice solution !
  $endgroup$
  – Yves Daoust
  Sep 20 at 8:37
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Elegant Solution . Really Beautiful and Beyond my Imaginationary Before I Discovered it. Thanks .
  $endgroup$
  – SAM.Am
  Sep 20 at 8:44
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @labbhattacharjee: good point! I have added a correction for that
  $endgroup$
  – robjohn♦
  Sep 20 at 8:48
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @robjohn, Thanks for the Iverson bracket which is new to me. Actually I was thinking in the same line as well.Regarding argument I think, we need en.wikipedia.org/wiki/Atan2#Definition_and_computation . For example, $x=dfrac17pi3$
  $endgroup$
  – lab bhattacharjee
  Sep 20 at 8:55
  

  
  
  
  

 | 
show 2 more comments

2

$begingroup$

$$1+e^ix=1+cos x+isin x=2cos^2frac x2+2icosfrac x2sinfrac x2=2cosfrac x2e^ix/2$$

so that

$$e^ix+e^2ix=2cosfrac x2 e^3ix/2.$$

---

Notice that the first locus is a shifted circle, while the second is the good old Pascal's limaçon, obtained by tripling the argument.

share|cite|improve this answer

edited Sep 20 at 8:40

answered Sep 20 at 8:33

Yves DaoustYves Daoust

151k1313 gold badges9191 silver badges251251 bronze badges

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  It's really a pleasure to read a such Gem . Thanks .
  $endgroup$
  – SAM.Am
  Sep 20 at 9:06
  

  
  
  
  

add a comment
 | 

1

$begingroup$

Let $$Z=e^ix+e^2ix=e^ix(1+e^ix) Rightarrow |Z|=|1+cos x +i sin x|= sqrt1+cos x)^2+sin ^2x$$ $$Rightarrow |Z|=sqrt2+2cos x=2 cos(x/2).$$

$$Arg(Z) =Arg[(cos x+cos 2x)+i(sin x+ sin 2x)]= tan^-1 fracsin x+sin 2xcos x + cos 2x$$ $$ Rightarrow Arg(Z)= tan^-1 tan (3x/2)=3x/2.$$

share|cite|improve this answer

answered Sep 20 at 8:24

Dr Zafar Ahmed DScDr Zafar Ahmed DSc

9,41811 gold badge44 silver badges2222 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Another piece of art . Elegant too.
  $endgroup$
  – SAM.Am
  Sep 20 at 8:49
  

  
  
  
  

add a comment
 | 

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5

$begingroup$

$$
beginalign
e^ix+e^2ix
&=e^frac32ixleft(e^frac12ix+e^-frac12ixright)\
&=e^icolor#090frac32xcolor#C002cosleft(tfrac12xright)
endalign
$$
So
$$
left|,e^ix+e^2ix,right|=left|,color#C002cosleft(tfrac12xright),right|
$$
and
$$
argleft(e^ix+e^2ixright)equivcolor#090tfrac32x+pileft[,cosleft(tfrac12xright)lt0,right]pmod2pi
$$
where $[cdots]$ are Iverson Brackets.

share|cite|improve this answer

edited Sep 20 at 9:14

answered Sep 20 at 8:36

robjohn♦robjohn

283k2929 gold badges335335 silver badges667667 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  But $$cosdfrac x2$$ can be $$<0?$$
  $endgroup$
  – lab bhattacharjee
  Sep 20 at 8:37
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Very nice solution !
  $endgroup$
  – Yves Daoust
  Sep 20 at 8:37
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Elegant Solution . Really Beautiful and Beyond my Imaginationary Before I Discovered it. Thanks .
  $endgroup$
  – SAM.Am
  Sep 20 at 8:44
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @labbhattacharjee: good point! I have added a correction for that
  $endgroup$
  – robjohn♦
  Sep 20 at 8:48
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @robjohn, Thanks for the Iverson bracket which is new to me. Actually I was thinking in the same line as well.Regarding argument I think, we need en.wikipedia.org/wiki/Atan2#Definition_and_computation . For example, $x=dfrac17pi3$
  $endgroup$
  – lab bhattacharjee
  Sep 20 at 8:55
  

  
  
  
  

 | 
show 2 more comments

5

$begingroup$

$$
beginalign
e^ix+e^2ix
&=e^frac32ixleft(e^frac12ix+e^-frac12ixright)\
&=e^icolor#090frac32xcolor#C002cosleft(tfrac12xright)
endalign
$$
So
$$
left|,e^ix+e^2ix,right|=left|,color#C002cosleft(tfrac12xright),right|
$$
and
$$
argleft(e^ix+e^2ixright)equivcolor#090tfrac32x+pileft[,cosleft(tfrac12xright)lt0,right]pmod2pi
$$
where $[cdots]$ are Iverson Brackets.

share|cite|improve this answer

edited Sep 20 at 9:14

answered Sep 20 at 8:36

robjohn♦robjohn

283k2929 gold badges335335 silver badges667667 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  But $$cosdfrac x2$$ can be $$<0?$$
  $endgroup$
  – lab bhattacharjee
  Sep 20 at 8:37
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Very nice solution !
  $endgroup$
  – Yves Daoust
  Sep 20 at 8:37
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Elegant Solution . Really Beautiful and Beyond my Imaginationary Before I Discovered it. Thanks .
  $endgroup$
  – SAM.Am
  Sep 20 at 8:44
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @labbhattacharjee: good point! I have added a correction for that
  $endgroup$
  – robjohn♦
  Sep 20 at 8:48
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @robjohn, Thanks for the Iverson bracket which is new to me. Actually I was thinking in the same line as well.Regarding argument I think, we need en.wikipedia.org/wiki/Atan2#Definition_and_computation . For example, $x=dfrac17pi3$
  $endgroup$
  – lab bhattacharjee
  Sep 20 at 8:55
  

  
  
  
  

 | 
show 2 more comments

$begingroup$

$$
beginalign
e^ix+e^2ix
&=e^frac32ixleft(e^frac12ix+e^-frac12ixright)\
&=e^icolor#090frac32xcolor#C002cosleft(tfrac12xright)
endalign
$$
So
$$
left|,e^ix+e^2ix,right|=left|,color#C002cosleft(tfrac12xright),right|
$$
and
$$
argleft(e^ix+e^2ixright)equivcolor#090tfrac32x+pileft[,cosleft(tfrac12xright)lt0,right]pmod2pi
$$
where $[cdots]$ are Iverson Brackets.

share|cite|improve this answer

edited Sep 20 at 9:14

answered Sep 20 at 8:36

robjohn♦robjohn

283k2929 gold badges335335 silver badges667667 bronze badges

$endgroup$

share|cite|improve this answer

edited Sep 20 at 9:14

answered Sep 20 at 8:36

robjohn♦robjohn

283k2929 gold badges335335 silver badges667667 bronze badges

answered Sep 20 at 8:36

robjohn♦robjohn

283k2929 gold badges335335 silver badges667667 bronze badges

answered Sep 20 at 8:36

robjohn♦robjohn

283k2929 gold badges335335 silver badges667667 bronze badges

2

$begingroup$

$$1+e^ix=1+cos x+isin x=2cos^2frac x2+2icosfrac x2sinfrac x2=2cosfrac x2e^ix/2$$

so that

$$e^ix+e^2ix=2cosfrac x2 e^3ix/2.$$

---

Notice that the first locus is a shifted circle, while the second is the good old Pascal's limaçon, obtained by tripling the argument.

share|cite|improve this answer

edited Sep 20 at 8:40

answered Sep 20 at 8:33

Yves DaoustYves Daoust

151k1313 gold badges9191 silver badges251251 bronze badges

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  It's really a pleasure to read a such Gem . Thanks .
  $endgroup$
  – SAM.Am
  Sep 20 at 9:06
  

  
  
  
  

add a comment
 | 

2

$begingroup$

$$1+e^ix=1+cos x+isin x=2cos^2frac x2+2icosfrac x2sinfrac x2=2cosfrac x2e^ix/2$$

so that

$$e^ix+e^2ix=2cosfrac x2 e^3ix/2.$$

---

Notice that the first locus is a shifted circle, while the second is the good old Pascal's limaçon, obtained by tripling the argument.

share|cite|improve this answer

edited Sep 20 at 8:40

answered Sep 20 at 8:33

Yves DaoustYves Daoust

151k1313 gold badges9191 silver badges251251 bronze badges

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  It's really a pleasure to read a such Gem . Thanks .
  $endgroup$
  – SAM.Am
  Sep 20 at 9:06
  

  
  
  
  

add a comment
 | 

$begingroup$

$$1+e^ix=1+cos x+isin x=2cos^2frac x2+2icosfrac x2sinfrac x2=2cosfrac x2e^ix/2$$

so that

$$e^ix+e^2ix=2cosfrac x2 e^3ix/2.$$

---

Notice that the first locus is a shifted circle, while the second is the good old Pascal's limaçon, obtained by tripling the argument.

share|cite|improve this answer

edited Sep 20 at 8:40

answered Sep 20 at 8:33

Yves DaoustYves Daoust

151k1313 gold badges9191 silver badges251251 bronze badges

$endgroup$

share|cite|improve this answer

edited Sep 20 at 8:40

answered Sep 20 at 8:33

Yves DaoustYves Daoust

151k1313 gold badges9191 silver badges251251 bronze badges

answered Sep 20 at 8:33

Yves DaoustYves Daoust

151k1313 gold badges9191 silver badges251251 bronze badges

answered Sep 20 at 8:33

Yves DaoustYves Daoust

151k1313 gold badges9191 silver badges251251 bronze badges

1

$begingroup$

Let $$Z=e^ix+e^2ix=e^ix(1+e^ix) Rightarrow |Z|=|1+cos x +i sin x|= sqrt1+cos x)^2+sin ^2x$$ $$Rightarrow |Z|=sqrt2+2cos x=2 cos(x/2).$$

$$Arg(Z) =Arg[(cos x+cos 2x)+i(sin x+ sin 2x)]= tan^-1 fracsin x+sin 2xcos x + cos 2x$$ $$ Rightarrow Arg(Z)= tan^-1 tan (3x/2)=3x/2.$$

share|cite|improve this answer

answered Sep 20 at 8:24

Dr Zafar Ahmed DScDr Zafar Ahmed DSc

9,41811 gold badge44 silver badges2222 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Another piece of art . Elegant too.
  $endgroup$
  – SAM.Am
  Sep 20 at 8:49
  

  
  
  
  

add a comment
 | 

1

$begingroup$

Let $$Z=e^ix+e^2ix=e^ix(1+e^ix) Rightarrow |Z|=|1+cos x +i sin x|= sqrt1+cos x)^2+sin ^2x$$ $$Rightarrow |Z|=sqrt2+2cos x=2 cos(x/2).$$

$$Arg(Z) =Arg[(cos x+cos 2x)+i(sin x+ sin 2x)]= tan^-1 fracsin x+sin 2xcos x + cos 2x$$ $$ Rightarrow Arg(Z)= tan^-1 tan (3x/2)=3x/2.$$

share|cite|improve this answer

answered Sep 20 at 8:24

Dr Zafar Ahmed DScDr Zafar Ahmed DSc

9,41811 gold badge44 silver badges2222 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Another piece of art . Elegant too.
  $endgroup$
  – SAM.Am
  Sep 20 at 8:49
  

  
  
  
  

add a comment
 | 

$begingroup$

Let $$Z=e^ix+e^2ix=e^ix(1+e^ix) Rightarrow |Z|=|1+cos x +i sin x|= sqrt1+cos x)^2+sin ^2x$$ $$Rightarrow |Z|=sqrt2+2cos x=2 cos(x/2).$$

$$Arg(Z) =Arg[(cos x+cos 2x)+i(sin x+ sin 2x)]= tan^-1 fracsin x+sin 2xcos x + cos 2x$$ $$ Rightarrow Arg(Z)= tan^-1 tan (3x/2)=3x/2.$$

share|cite|improve this answer

answered Sep 20 at 8:24

Dr Zafar Ahmed DScDr Zafar Ahmed DSc

9,41811 gold badge44 silver badges2222 bronze badges

$endgroup$

share|cite|improve this answer

answered Sep 20 at 8:24

Dr Zafar Ahmed DScDr Zafar Ahmed DSc

9,41811 gold badge44 silver badges2222 bronze badges

answered Sep 20 at 8:24

Dr Zafar Ahmed DScDr Zafar Ahmed DSc

9,41811 gold badge44 silver badges2222 bronze badges

answered Sep 20 at 8:24

Dr Zafar Ahmed DScDr Zafar Ahmed DSc

9,41811 gold badge44 silver badges2222 bronze badges