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I am puzzled by this function:

f(x)= 139.85 + (15.8404 + 4.76022 I) E^(-I x) + (15.8404 - 4.76022 I) E^( I x)

It seems to have only an imaginary part. However, its first half:

g(x)= 139.85 + (15.8404 + 4.76022 I) E^(-I x)

has both an imaginary and a real part.

What is the explanation for this?

Is the plotting command used wrong?

Plot[Evaluate[ReIm[139.85 + (15.8404 + 4.76022 I)/E^(I x) + (15.8404 - 4.76022 I) E^(I x)]], {x, -6., 6.}]
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  • $\begingroup$ There's no reason to think that a sum must have a nonzero imaginary part if one piece of it does. A similar fallacy would be "1.24+5.76 has only an integer part but its first half 1.24 has both an integer and a decimal part". $\endgroup$
    – Greg Martin
    Mar 9 at 0:11

3 Answers 3

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Take notice that your function is of the form:

g = c + (a + b I) E^(-I x) + (a - b I) E^(I x) // FullSimplify

enter image description here

which only has real components.

ReIm makes assumptions about x and treats it as a complex number itself. You can use ComplexExpand to solve your issue (and put your mind at ease that it is doing the right thing :) )

ComplexExpand[
 ReIm[139.85 + (15.8404 + 4.76022 I)/
    E^(I x) + (15.8404 - 4.76022 I) E^(I x)]]

enter image description here

As you can see, it only gives a real part.

Best wishes!

A.

Edit: I really liked @a gues answer which helps to think more geometrically. I'm only adding the following for completion.

z = a + b I /. {a -> 1, b -> 1};
z2 = z E^(-I \[Pi]/12.);
ListPlot[{{ReIm[z], ReIm[z\[Conjugate]], 
   ReIm[z + z\[Conjugate]]}, {ReIm[z2], ReIm[z2\[Conjugate]], 
   ReIm[z2] + ReIm[z2\[Conjugate]]}}, 
 Epilog -> { Arrow[{{0, 0}, ReIm[z]}], 
   Arrow[{{0, 0}, ReIm[z\[Conjugate]]}], Green, 
   Arrow[{{0, 0}, ReIm[z + z\[Conjugate]]}], Orange,
   Arrow[{{0, 0}, ReIm[z2]}], Arrow[{{0, 0}, ReIm[z2\[Conjugate]]}], 
   Red, Arrow[{{0, 0}, ReIm[z2 + z2\[Conjugate]]}], Purple, Dashed, 
   Arrow[BezierCurve[{ReIm[z], ReIm[(z + z2)/1.5], ReIm[z2]}]], 
   Arrow[BezierCurve[{ReIm[z\[Conjugate]], 
      ReIm[(z\[Conjugate] + z2\[Conjugate])/1.5], 
      ReIm[z2\[Conjugate]]}]]}, PlotRange -> {{0, 3}, Full}]

enter image description here

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Your function

f[x_] := 139.85 + (15.8404 + 4.76022 I)/E^(I x) + (15.8404 - 4.76022 I) E^(I x)

is real valued, because definition

fR[x_] := 139.85 + (15.8404 + 4.76022 I)/E^(I x) + Conjugate[(15.8404 + 4.76022 I)/E^(I x)]
Simplify[f[x] == fR[x], Element[x, Reals]]
(* True *)

is equivalent for real x.

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you are basically rotating to vectors in opposite directions and wondering why their sum stays on the line that bijects them, because the rotations are equal and opposite their imaginary components will also always be equal and opposite.

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  • $\begingroup$ I like your more visual approach of thinking about it. It's easier to simply do the math and not think about it. I've edited my post to better depict your idea. $\endgroup$
    – alex
    Mar 8 at 21:47

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