# How to express the given function as a factor $e^{7 i x}$ times a real function?

I have the function $$f(x,b,t,m)$$ for $$\{x>0,t,b\}$$ all reals and $$m=\{1,2,3,4,5,6\}$$. For each particular $$m\in\{1,2,3,4,5,6\}$$, I see that $$f(x,b,t,m)$$ has a multiplicative factor $$e^{7 i x}$$ times another real function, for example, for $$m=1$$, we have

$$e^{7 i x} \left(\cos 7 b+\cos t+448 \cos 5 x+128 \cos 7 x+14 \cos x)(251-342 \sin \frac{\pi }{14}+326 \sin \frac{3 \pi }{14}-346 \cos \frac{\pi }{7})-112 \cos 3 x(17 \sin \frac{\pi }{14}-16 (1+\sin \frac{3 \pi }{14})+17 \cos \frac{\pi }{7})\right)$$

I want to get a similar result for the general $$m$$; in other words, to show that $$f(x,b,t,m)$$ can be expressed in terms of this multiplicative factor $$e^{7 i x}$$ times another real function which is $$m$$ dependent.

f[x_,b_,t_,m_]:=64 + 336 E^(2 I x) + 1008 E^(4 I x) + 1757 E^(6 I x) +
1757 E^(8 I x) + 1008 E^(10 I x) + 336 E^(12 I x) + 64 E^(14 I x) +
E^(7 I x) Cos[7 b] +
8 E^(2 I x) (24 + 110 E^(2 I x) + 215 E^(4 I x) + 215 E^(6 I x) +
110 E^(8 I x) + 12 E^(10 I x)) Cos[(2 m \[Pi])/7] +
4 E^(2 I x) (40 + 202 E^(2 I x) + 201 E^(4 I x) + 201 E^(6 I x) +
202 E^(8 I x) + 40 E^(10 I x)) Cos[(4 m \[Pi])/7] +
4 E^(2 I x) (32 + 82 E^(2 I x) + 335 E^(4 I x) + 335 E^(6 I x) +
82 E^(8 I x) + 32 E^(10 I x)) Cos[(6 m \[Pi])/7] +
2 E^(2 I x) (48 + 128 E^(2 I x) + 527 E^(4 I x) + 527 E^(6 I x) +
128 E^(8 I x) + 48 E^(10 I x)) Cos[(8 m \[Pi])/7] +
2 E^(2 I x) (32 + 176 E^(2 I x) + 183 E^(4 I x) + 183 E^(6 I x) +
176 E^(8 I x) + 32 E^(10 I x)) Cos[(10 m \[Pi])/7] +
4 E^(2 I x) (8 + 50 E^(2 I x) + 111 E^(4 I x) + 111 E^(6 I x) +
50 E^(8 I x) + 32 E^(10 I x)) Cos[(12 m \[Pi])/7] +
2 E^(4 I x) (20 + 59 E^(2 I x) + 59 E^(4 I x) + 20 E^(6 I x)) Cos[(
16 m \[Pi])/7] +
2 E^(4 I x) (8 + 13 E^(2 I x) + 13 E^(4 I x) + 8 E^(6 I x)) Cos[(
18 m \[Pi])/7] +
8 E^(4 I x) (74 + 3 E^(2 I x) + 3 E^(4 I x) + 74 E^(6 I x)) Cos[(
20 m \[Pi])/7] + 4 E^(6 I x) (1 + E^(2 I x)) Cos[(22 m \[Pi])/7] +
1198 E^(6 I x) (1 + E^(2 I x)) Cos[(24 m \[Pi])/7] +
E^(7 I x) Cos[t] - 96 I E^(12 I x) Sin[(2 m \[Pi])/7] -
804 I E^(6 I x) (1 + E^(2 I x)) Sin[(4 m \[Pi])/7] +
328 I E^(4 I x) (1 + E^(6 I x)) Sin[(6 m \[Pi])/7] -
256 I E^(4 I x) (1 + E^(6 I x)) Sin[(8 m \[Pi])/7] +
366 I E^(6 I x) (1 + E^(2 I x)) Sin[(10 m \[Pi])/7] -
96 I E^(12 I x) Sin[(12 m \[Pi])/7] -
26 I E^(6 I x) (1 + E^(2 I x)) Sin[(18 m \[Pi])/7] -
584 I E^(4 I x) (1 + E^(6 I x)) Sin[(20 m \[Pi])/7] -
1196 I E^(6 I x) (1 + E^(2 I x)) Sin[(24 m \[Pi])/7]


• Sounds like you want $f(x,b,t,m)/e^{7 i x}$, since the function is $e^{7 i x}$ times this factor; but you probably mean something else. Commented Oct 29, 2022 at 1:42
• @MichaelE2 Thanks for your comment. No, the problem is exactly this; I want to show that for a general $m$, the function $f(x,b,t,m)$ has (as it must) this coefficient $e^{7ix}$. I tried FullSimplify for the general case, but maybe my computer is slow and it takes a lot of time. I thought maybe there is a simple way to simplify the function. Commented Oct 29, 2022 at 1:59
• But it seems to be what I said then, no? For instance f[x, b, t, m]/E^(7 I x) // ReIm // ComplexExpand // {1, I}.Simplify[#, m \[Element] Integers && 1 <= m <= 6] & Commented Oct 29, 2022 at 2:30

$Version (* "13.1.0 for Mac OS X x86 (64-bit) (June 16, 2022)" *) Clear["Global*"] f[x_, b_, t_, m_] := 64 + 336 E^(2 I x) + 1008 E^(4 I x) + 1757 E^(6 I x) + 1757 E^(8 I x) + 1008 E^(10 I x) + 336 E^(12 I x) + 64 E^(14 I x) + E^(7 I x) Cos[7 b] + 8 E^(2 I x) (24 + 110 E^(2 I x) + 215 E^(4 I x) + 215 E^(6 I x) + 110 E^(8 I x) + 12 E^(10 I x)) Cos[(2 m π)/7] + 4 E^(2 I x) (40 + 202 E^(2 I x) + 201 E^(4 I x) + 201 E^(6 I x) + 202 E^(8 I x) + 40 E^(10 I x)) Cos[(4 m π)/7] + 4 E^(2 I x) (32 + 82 E^(2 I x) + 335 E^(4 I x) + 335 E^(6 I x) + 82 E^(8 I x) + 32 E^(10 I x)) Cos[(6 m π)/7] + 2 E^(2 I x) (48 + 128 E^(2 I x) + 527 E^(4 I x) + 527 E^(6 I x) + 128 E^(8 I x) + 48 E^(10 I x)) Cos[(8 m π)/7] + 2 E^(2 I x) (32 + 176 E^(2 I x) + 183 E^(4 I x) + 183 E^(6 I x) + 176 E^(8 I x) + 32 E^(10 I x)) Cos[(10 m π)/7] + 4 E^(2 I x) (8 + 50 E^(2 I x) + 111 E^(4 I x) + 111 E^(6 I x) + 50 E^(8 I x) + 32 E^(10 I x)) Cos[(12 m π)/7] + 2 E^(4 I x) (20 + 59 E^(2 I x) + 59 E^(4 I x) + 20 E^(6 I x)) Cos[(16 m π)/7] + 2 E^(4 I x) (8 + 13 E^(2 I x) + 13 E^(4 I x) + 8 E^(6 I x)) Cos[(18 m π)/7] + 8 E^(4 I x) (74 + 3 E^(2 I x) + 3 E^(4 I x) + 74 E^(6 I x)) Cos[(20 m π)/7] + 4 E^(6 I x) (1 + E^(2 I x)) Cos[(22 m π)/7] + 1198 E^(6 I x) (1 + E^(2 I x)) Cos[(24 m π)/7] + E^(7 I x) Cos[t] - 96 I E^(12 I x) Sin[(2 m π)/7] - 804 I E^(6 I x) (1 + E^(2 I x)) Sin[(4 m π)/7] + 328 I E^(4 I x) (1 + E^(6 I x)) Sin[(6 m π)/7] - 256 I E^(4 I x) (1 + E^(6 I x)) Sin[(8 m π)/7] + 366 I E^(6 I x) (1 + E^(2 I x)) Sin[(10 m π)/7] - 96 I E^(12 I x) Sin[(12 m π)/7] - 26 I E^(6 I x) (1 + E^(2 I x)) Sin[(18 m π)/7] - 584 I E^(4 I x) (1 + E^(6 I x)) Sin[(20 m π)/7] - 1196 I E^(6 I x) (1 + E^(2 I x)) Sin[(24 m π)/7];  Restructuring f (expr = f[x, b, t, m] // ComplexExpand // Simplify) // Short[#, 2] &  The initial factor of expr is E^(7*I*x) (expr2 = ReplacePart[expr, 1 -> TrigToExp[expr[[1]]]]) // Short  Verifying that expr2 is equivalent to f[x, b, t, m] f[x, b, t, m] == expr2 // Simplify (* True *)  • Thanks. I need the complete form of the function in your brackets (which should be$m\$ dependent); can we obtain it? Commented Oct 29, 2022 at 2:10
• The expressions are complete. Short is used to suppress their full display (note that the parentheses are used to isolate the Short from the expressions definitions). Remove each Short` and you will see the full expressions. Commented Oct 29, 2022 at 2:17