Menu Close

f-C-C-a-b-R-2-a-lt-b-f-z-a-f-b-z-sin-zpi-b-a-f-z-z-2-z-a-b-2-f-z-0-z-




Question Number 2362 by 123456 last updated on 18/Nov/15
f:C→C,(a,b)∈R^2 ,a<b  f(z−a)=f(b−z)sin ((zπ)/(b−a))  f(z)=z^2 ,ℜ(z)≥((a+b)/2)  f(z)=0,z=?
$${f}:\mathbb{C}\rightarrow\mathbb{C},\left({a},{b}\right)\in\mathbb{R}^{\mathrm{2}} ,{a}<{b} \\ $$$${f}\left({z}−{a}\right)={f}\left({b}−{z}\right)\mathrm{sin}\:\frac{{z}\pi}{{b}−{a}} \\ $$$${f}\left({z}\right)={z}^{\mathrm{2}} ,\Re\left({z}\right)\geqslant\frac{{a}+{b}}{\mathrm{2}} \\ $$$${f}\left({z}\right)=\mathrm{0},{z}=? \\ $$
Commented by Rasheed Soomro last updated on 20/Nov/15
Let a=0 then b>0:  f(z−a)=f(b−z)sin ((zπ)/(b−a))⇒f(z)=f(b−z)sin ((zπ)/b)=0 if f(z)=0  ⇒f(b−z)=0 ∣{ sin ((zπ)/b)=0⇒((zπ)/b)=sin^(−1) 0⇒z=0}  f(z)=0⇒z=0 is one solution.  For a=0 and b>0       f(z)=f(b−z)sin ((zπ)/b).............(i)       f(b−z)=f(b−b−z^(−) )sin(((b−z)π)/b)                           =f(z)sin(((b−z)π)/b).......(ii)  From (i) and (ii)      f(z)=(f(z)sin(((b−z)π)/b))sin ((zπ)/b)      (sin(((b−z)π)/b))(sin ((zπ)/b))=1      (1/2){cos((((b−z)π)/b)−((zπ)/b))−cos((((b−z)π)/b)+((zπ)/b))}=1      (1/2){cos((((b−z−z)π)/b))−cos((((b−z+z)π)/b))}=1      cos((((b−2z)π)/b))−cos((π)=2      cos((((b−2z)π)/b))+1=2      cos((((b−2z)π)/b))=1     (((b−2z)π)/b)=cos^(−1) (1)=0     b−2z=0          z=(b/2)  Continue
$${Let}\:{a}=\mathrm{0}\:{then}\:{b}>\mathrm{0}: \\ $$$${f}\left({z}−{a}\right)={f}\left({b}−{z}\right)\mathrm{sin}\:\frac{{z}\pi}{{b}−{a}}\Rightarrow{f}\left({z}\right)={f}\left({b}−{z}\right)\mathrm{sin}\:\frac{{z}\pi}{{b}}=\mathrm{0}\:{if}\:{f}\left({z}\right)=\mathrm{0} \\ $$$$\Rightarrow{f}\left({b}−{z}\right)=\mathrm{0}\:\mid\left\{\:\mathrm{sin}\:\frac{{z}\pi}{{b}}=\mathrm{0}\Rightarrow\frac{{z}\pi}{{b}}={sin}^{−\mathrm{1}} \mathrm{0}\Rightarrow{z}=\mathrm{0}\right\} \\ $$$${f}\left({z}\right)=\mathrm{0}\Rightarrow{z}=\mathrm{0}\:{is}\:{one}\:{solution}. \\ $$$${For}\:{a}=\mathrm{0}\:{and}\:{b}>\mathrm{0} \\ $$$$\:\:\:\:\:{f}\left({z}\right)={f}\left({b}−{z}\right)\mathrm{sin}\:\frac{{z}\pi}{{b}}………….\left({i}\right) \\ $$$$\:\:\:\:\:{f}\left({b}−{z}\right)={f}\left({b}−\overline {{b}−{z}}\right){sin}\frac{\left({b}−{z}\right)\pi}{{b}}\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:={f}\left({z}\right){sin}\frac{\left({b}−{z}\right)\pi}{{b}}…….\left({ii}\right) \\ $$$${From}\:\left({i}\right)\:{and}\:\left({ii}\right) \\ $$$$\:\:\:\:{f}\left({z}\right)=\left({f}\left({z}\right){sin}\frac{\left({b}−{z}\right)\pi}{{b}}\right)\mathrm{sin}\:\frac{{z}\pi}{{b}} \\ $$$$\:\:\:\:\left({sin}\frac{\left({b}−{z}\right)\pi}{{b}}\right)\left(\mathrm{sin}\:\frac{{z}\pi}{{b}}\right)=\mathrm{1} \\ $$$$\:\:\:\:\frac{\mathrm{1}}{\mathrm{2}}\left\{{cos}\left(\frac{\left({b}−{z}\right)\pi}{{b}}−\frac{{z}\pi}{{b}}\right)−{cos}\left(\frac{\left({b}−{z}\right)\pi}{{b}}+\frac{{z}\pi}{{b}}\right)\right\}=\mathrm{1} \\ $$$$\:\:\:\:\frac{\mathrm{1}}{\mathrm{2}}\left\{{cos}\left(\frac{\left({b}−{z}−{z}\right)\pi}{{b}}\right)−{cos}\left(\frac{\left({b}−{z}+{z}\right)\pi}{{b}}\right)\right\}=\mathrm{1} \\ $$$$\:\:\:\:{cos}\left(\frac{\left({b}−\mathrm{2}{z}\right)\pi}{{b}}\right)−{cos}\left(\left(\pi\right)=\mathrm{2}\right. \\ $$$$\:\:\:\:{cos}\left(\frac{\left({b}−\mathrm{2}{z}\right)\pi}{{b}}\right)+\mathrm{1}=\mathrm{2} \\ $$$$\:\:\:\:{cos}\left(\frac{\left({b}−\mathrm{2}{z}\right)\pi}{{b}}\right)=\mathrm{1} \\ $$$$\:\:\:\frac{\left({b}−\mathrm{2}{z}\right)\pi}{{b}}={cos}^{−\mathrm{1}} \left(\mathrm{1}\right)=\mathrm{0} \\ $$$$\:\:\:{b}−\mathrm{2}{z}=\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:{z}=\frac{{b}}{\mathrm{2}} \\ $$$$\mathcal{C}{ontinue} \\ $$$$\:\:\:\:\: \\ $$
Commented by Rasheed Soomro last updated on 20/Nov/15
The condition ℜ(z)≥((a+b)/2)  is only for f(z)=z^2   ?
$${The}\:{condition}\:\Re\left({z}\right)\geqslant\frac{{a}+{b}}{\mathrm{2}}\:\:{is}\:{only}\:{for}\:{f}\left({z}\right)={z}^{\mathrm{2}} \:\:? \\ $$
Commented by Rasheed Soomro last updated on 21/Nov/15
When you write statement and condition on   one line it should mean that the condition is  only for that statement.For example the condition  ℜ(z)≥((a+b)/2) is only in case f(z)=z^2  :  f(z)=z^2 ,ℜ(z)≥((a+b)/2)  Similarly in the following you are asked for z when  f(z)=0.  :  f(z)=0,z=?  Am I Right?
$${When}\:{you}\:{write}\:{statement}\:{and}\:{condition}\:{on}\: \\ $$$${one}\:{line}\:{it}\:{should}\:{mean}\:{that}\:{the}\:{condition}\:{is} \\ $$$${only}\:{for}\:{that}\:{statement}.{For}\:{example}\:{the}\:{condition} \\ $$$$\Re\left({z}\right)\geqslant\frac{{a}+{b}}{\mathrm{2}}\:{is}\:{only}\:{in}\:{case}\:{f}\left({z}\right)={z}^{\mathrm{2}} \:: \\ $$$${f}\left({z}\right)={z}^{\mathrm{2}} ,\Re\left({z}\right)\geqslant\frac{{a}+{b}}{\mathrm{2}} \\ $$$${Similarly}\:{in}\:{the}\:{following}\:{you}\:{are}\:{asked}\:{for}\:{z}\:{when} \\ $$$${f}\left({z}\right)=\mathrm{0}.\:\:: \\ $$$${f}\left({z}\right)=\mathrm{0},{z}=? \\ $$$$\mathcal{A}{m}\:\mathcal{I}\:\mathcal{R}{ight}? \\ $$
Answered by Rasheed Soomro last updated on 21/Nov/15
f:C→C,(a,b)∈R^2 ,a<b  f(z−a)=f(b−z)sin ((zπ)/(b−a))  f(z)=z^2 ,ℜ(z)≥((a+b)/2)  f(z)=0,z=?  −−−−−−−−−−−−−−−−−−−−−  f(z−a)=f(b−z)sin ((zπ)/(b−a)) is true for any real a,b with a<b  Let a=0,then b>0  f(z−a)=f(b−z)sin ((zπ)/(b−a))          ⇒f(z)=f(b−z)sin ((zπ)/b)            ⇒ f(b−z)=f(b−b−z^(−) )sin ((zπ)/b)                                   =f(z)sin ((zπ)/b)  f(z)=f(z)sin ((zπ)/b)sin ((zπ)/b)  f(z)−f(z)sin^2  ((zπ)/b)=0  f(z)(1−sin^2  ((zπ)/b))=0  f(z)(1−sin((zπ)/b))(1+sin((zπ)/b))=0  f(z)=0   ∣     sin((πz)/b) =1  ∣   sin ((πz)/b)=−1                    ∣  ((πz)/b)=sin^(−1) (1)  ∣  ((πz)/b)=sin^(−1) (−1)                    ∣ ((πz)/b)=(π/2)  ∣  ((πz)/b)=((3π)/2)                    ∣ z=(b/2)     ∣ z=((3b)/2)        for a=0 ,b>0                    ∣  ℜ(z)≥ ((a+b)/2)=((0+b)/2)                    ∣  (b/2)≥(b/2)      Condition satisfied for z=(b/2)                    ∣ ((3b)/2)≥(b/2)        as well as    for z=((3b)/2)  −−−−−−−−−−−−−  Let b=0 [a<0]  f(z−a)=f(b−z)sin ((zπ)/(b−a))  ⇒f(z−a)=f(−z)sin ((zπ)/(−a))  ⇒f(z−a)=−f(−z)sin ((zπ)/a)  f(−z) =−((f(z−a))/(sin ((zπ)/a)))          f(z−a)=f(−a−z^(−) )=−((f(a−z−a))/(sin (((a−z)π)/a)))                                           =−((f(−z))/(sin (((a−z)π)/a)))  f(−z) =−((−((f(−z))/(sin (((a−z)π)/a))))/(sin ((zπ)/a)))  f(−z)=((f(−z))/(sin (((a−z)π)/a).sin ((zπ)/a)))  f(−z)−((f(−z))/(sin (((a−z)π)/a).sin ((zπ)/a)))=0  f(−z)[1−(1/(sin (((a−z)π)/a).sin ((zπ)/a)))]=0  f(−z)=0  ∣  (1/(sin (((a−z)π)/a).sin ((zπ)/a)))=1                       ∣  sin (((a−z)π)/a).sin ((zπ)/a)=1                       ∣ (1/2){cos((((a−z−z)π)/a))−cos((((a−z+z)π)/a)}=1                       ∣ (1/2){cos((((a−2z)π)/a))−cos((((a)π)/a)}=1                       ∣ (1/2){cos((((a−2z)π)/a))−cos(π)}=1                       ∣ cos((((a−2z)π)/a))+1=2                       ∣ cos((((a−2z)π)/a))=1                       ∣   (((a−2z)π)/a)=cos^(−1) (1)=0                       ∣  a−2z=0⇒z=(a/2)     [Satisfy the condition ℜ(z)≥((a+b)/2)] [b=0
$${f}:\mathbb{C}\rightarrow\mathbb{C},\left({a},{b}\right)\in\mathbb{R}^{\mathrm{2}} ,{a}<{b} \\ $$$${f}\left({z}−{a}\right)={f}\left({b}−{z}\right)\mathrm{sin}\:\frac{{z}\pi}{{b}−{a}} \\ $$$${f}\left({z}\right)={z}^{\mathrm{2}} ,\Re\left({z}\right)\geqslant\frac{{a}+{b}}{\mathrm{2}} \\ $$$${f}\left({z}\right)=\mathrm{0},{z}=? \\ $$$$−−−−−−−−−−−−−−−−−−−−− \\ $$$${f}\left({z}−{a}\right)={f}\left({b}−{z}\right)\mathrm{sin}\:\frac{{z}\pi}{{b}−{a}}\:{is}\:{true}\:{for}\:{any}\:{real}\:{a},{b}\:{with}\:{a}<{b} \\ $$$${Let}\:{a}=\mathrm{0},{then}\:{b}>\mathrm{0} \\ $$$${f}\left({z}−{a}\right)={f}\left({b}−{z}\right)\mathrm{sin}\:\frac{{z}\pi}{{b}−{a}} \\ $$$$\:\:\:\:\:\:\:\:\Rightarrow{f}\left({z}\right)={f}\left({b}−{z}\right)\mathrm{sin}\:\frac{{z}\pi}{{b}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\Rightarrow\:{f}\left({b}−{z}\right)={f}\left({b}−\overline {{b}−{z}}\right)\mathrm{sin}\:\frac{{z}\pi}{{b}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:={f}\left({z}\right)\mathrm{sin}\:\frac{{z}\pi}{{b}} \\ $$$${f}\left({z}\right)={f}\left({z}\right)\mathrm{sin}\:\frac{{z}\pi}{{b}}\mathrm{sin}\:\frac{{z}\pi}{{b}} \\ $$$${f}\left({z}\right)−{f}\left({z}\right)\mathrm{sin}^{\mathrm{2}} \:\frac{{z}\pi}{{b}}=\mathrm{0} \\ $$$${f}\left({z}\right)\left(\mathrm{1}−\mathrm{sin}^{\mathrm{2}} \:\frac{{z}\pi}{{b}}\right)=\mathrm{0} \\ $$$${f}\left({z}\right)\left(\mathrm{1}−\mathrm{sin}\frac{{z}\pi}{{b}}\right)\left(\mathrm{1}+\mathrm{sin}\frac{{z}\pi}{{b}}\right)=\mathrm{0} \\ $$$${f}\left({z}\right)=\mathrm{0}\:\:\:\mid\:\:\:\:\:\mathrm{sin}\frac{\pi{z}}{{b}}\:=\mathrm{1}\:\:\mid\:\:\:\mathrm{sin}\:\frac{\pi{z}}{{b}}=−\mathrm{1} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mid\:\:\frac{\pi{z}}{{b}}=\mathrm{sin}^{−\mathrm{1}} \left(\mathrm{1}\right)\:\:\mid\:\:\frac{\pi{z}}{{b}}=\mathrm{sin}^{−\mathrm{1}} \left(−\mathrm{1}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mid\:\frac{\pi{z}}{{b}}=\frac{\pi}{\mathrm{2}}\:\:\mid\:\:\frac{\pi{z}}{{b}}=\frac{\mathrm{3}\pi}{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mid\:{z}=\frac{{b}}{\mathrm{2}}\:\:\:\:\:\mid\:{z}=\frac{\mathrm{3}{b}}{\mathrm{2}}\:\:\:\:\:\:\:\:{for}\:{a}=\mathrm{0}\:,{b}>\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mid\:\:\Re\left({z}\right)\geqslant\:\frac{{a}+{b}}{\mathrm{2}}=\frac{\mathrm{0}+{b}}{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mid\:\:\frac{{b}}{\mathrm{2}}\geqslant\frac{{b}}{\mathrm{2}}\:\:\:\:\:\:{Condition}\:{satisfied}\:{for}\:{z}=\frac{{b}}{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mid\:\frac{\mathrm{3}{b}}{\mathrm{2}}\geqslant\frac{{b}}{\mathrm{2}}\:\:\:\:\:\:\:\:{as}\:{well}\:{as}\:\:\:\:{for}\:{z}=\frac{\mathrm{3}{b}}{\mathrm{2}} \\ $$$$−−−−−−−−−−−−− \\ $$$${Let}\:{b}=\mathrm{0}\:\left[{a}<\mathrm{0}\right] \\ $$$${f}\left({z}−{a}\right)={f}\left({b}−{z}\right)\mathrm{sin}\:\frac{{z}\pi}{{b}−{a}} \\ $$$$\Rightarrow{f}\left({z}−{a}\right)={f}\left(−{z}\right)\mathrm{sin}\:\frac{{z}\pi}{−{a}} \\ $$$$\Rightarrow{f}\left({z}−{a}\right)=−{f}\left(−{z}\right)\mathrm{sin}\:\frac{{z}\pi}{{a}} \\ $$$${f}\left(−{z}\right)\:=−\frac{{f}\left({z}−{a}\right)}{\mathrm{sin}\:\frac{{z}\pi}{{a}}} \\ $$$$\:\:\:\:\:\:\:\:{f}\left({z}−{a}\right)={f}\left(−\overline {{a}−{z}}\right)=−\frac{{f}\left({a}−{z}−{a}\right)}{\mathrm{sin}\:\frac{\left({a}−{z}\right)\pi}{{a}}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=−\frac{{f}\left(−{z}\right)}{\mathrm{sin}\:\frac{\left({a}−{z}\right)\pi}{{a}}} \\ $$$${f}\left(−{z}\right)\:=−\frac{−\frac{{f}\left(−{z}\right)}{\mathrm{sin}\:\frac{\left({a}−{z}\right)\pi}{{a}}}}{\mathrm{sin}\:\frac{{z}\pi}{{a}}} \\ $$$${f}\left(−{z}\right)=\frac{{f}\left(−{z}\right)}{\mathrm{sin}\:\frac{\left({a}−{z}\right)\pi}{{a}}.\mathrm{sin}\:\frac{{z}\pi}{{a}}} \\ $$$${f}\left(−{z}\right)−\frac{{f}\left(−{z}\right)}{\mathrm{sin}\:\frac{\left({a}−{z}\right)\pi}{{a}}.\mathrm{sin}\:\frac{{z}\pi}{{a}}}=\mathrm{0} \\ $$$${f}\left(−{z}\right)\left[\mathrm{1}−\frac{\mathrm{1}}{\mathrm{sin}\:\frac{\left({a}−{z}\right)\pi}{{a}}.\mathrm{sin}\:\frac{{z}\pi}{{a}}}\right]=\mathrm{0} \\ $$$${f}\left(−{z}\right)=\mathrm{0}\:\:\mid\:\:\frac{\mathrm{1}}{\mathrm{sin}\:\frac{\left({a}−{z}\right)\pi}{{a}}.\mathrm{sin}\:\frac{{z}\pi}{{a}}}=\mathrm{1} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mid\:\:\mathrm{sin}\:\frac{\left({a}−{z}\right)\pi}{{a}}.\mathrm{sin}\:\frac{{z}\pi}{{a}}=\mathrm{1} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mid\:\frac{\mathrm{1}}{\mathrm{2}}\left\{{cos}\left(\frac{\left({a}−{z}−{z}\right)\pi}{{a}}\right)−{cos}\left(\frac{\left({a}−{z}+{z}\right)\pi}{{a}}\right\}=\mathrm{1}\right. \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mid\:\frac{\mathrm{1}}{\mathrm{2}}\left\{{cos}\left(\frac{\left({a}−\mathrm{2}{z}\right)\pi}{{a}}\right)−{cos}\left(\frac{\left({a}\right)\pi}{{a}}\right\}=\mathrm{1}\right. \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mid\:\frac{\mathrm{1}}{\mathrm{2}}\left\{{cos}\left(\frac{\left({a}−\mathrm{2}{z}\right)\pi}{{a}}\right)−{cos}\left(\pi\right)\right\}=\mathrm{1} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mid\:{cos}\left(\frac{\left({a}−\mathrm{2}{z}\right)\pi}{{a}}\right)+\mathrm{1}=\mathrm{2} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mid\:{cos}\left(\frac{\left({a}−\mathrm{2}{z}\right)\pi}{{a}}\right)=\mathrm{1} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mid\:\:\:\frac{\left({a}−\mathrm{2}{z}\right)\pi}{{a}}={cos}^{−\mathrm{1}} \left(\mathrm{1}\right)=\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mid\:\:{a}−\mathrm{2}{z}=\mathrm{0}\Rightarrow{z}=\frac{{a}}{\mathrm{2}}\:\:\:\:\:\left[{Satisfy}\:{the}\:{condition}\:\Re\left({z}\right)\geqslant\frac{{a}+{b}}{\mathrm{2}}\right]\:\left[{b}=\mathrm{0}\right. \\ $$$$ \\ $$
Commented by Rasheed Soomro last updated on 21/Nov/15
This is not full answer.It is for a=0 or b=0 and also  z is pure real.  Needs verification  and  attestation.
$${This}\:{is}\:{not}\:{full}\:{answer}.{It}\:{is}\:{for}\:{a}=\mathrm{0}\:{or}\:{b}=\mathrm{0}\:{and}\:{also} \\ $$$${z}\:{is}\:{pure}\:{real}. \\ $$$${Needs}\:{verification}\:\:{and}\:\:{attestation}. \\ $$
Answered by Rasheed Soomro last updated on 22/Nov/15
f:C→C,(a,b)∈R^2 ,a<b  f(z−a)=f(b−z)sin ((zπ)/(b−a))  f(z)=z^2 ,ℜ(z)≥((a+b)/2)  f(z)=0,z=?  −−−−−−−−−−−−−−−−−−−−−  Let z=x+iy  f(z−a)=f(b−z)sin ((zπ)/(b−a))⇒f((x+iy)−a)=f(b−(x+iy))sin (((x+iy)π)/(b−a))  Let  b−a=1 then b=a+1    f((x+iy)−a)=f( a+1−(x+iy) )sin (x+iy)  Continue
$${f}:\mathbb{C}\rightarrow\mathbb{C},\left({a},{b}\right)\in\mathbb{R}^{\mathrm{2}} ,{a}<{b} \\ $$$${f}\left({z}−{a}\right)={f}\left({b}−{z}\right)\mathrm{sin}\:\frac{{z}\pi}{{b}−{a}} \\ $$$${f}\left({z}\right)={z}^{\mathrm{2}} ,\Re\left({z}\right)\geqslant\frac{{a}+{b}}{\mathrm{2}} \\ $$$${f}\left({z}\right)=\mathrm{0},{z}=? \\ $$$$−−−−−−−−−−−−−−−−−−−−− \\ $$$${Let}\:{z}={x}+{iy} \\ $$$${f}\left({z}−{a}\right)={f}\left({b}−{z}\right)\mathrm{sin}\:\frac{{z}\pi}{{b}−{a}}\Rightarrow{f}\left(\left({x}+{iy}\right)−{a}\right)={f}\left({b}−\left({x}+{iy}\right)\right)\mathrm{sin}\:\frac{\left({x}+{iy}\right)\pi}{{b}−{a}} \\ $$$${Let}\:\:{b}−{a}=\mathrm{1}\:{then}\:{b}={a}+\mathrm{1} \\ $$$$\:\:{f}\left(\left({x}+{iy}\right)−{a}\right)={f}\left(\:{a}+\mathrm{1}−\left({x}+{iy}\right)\:\right)\mathrm{sin}\:\left({x}+{iy}\right) \\ $$$${Continue} \\ $$

Leave a Reply

Your email address will not be published. Required fields are marked *