Question Number 2362 by 123456 last updated on 18/Nov/15
$${f}:\mathbb{C}\rightarrow\mathbb{C},\left({a},{b}\right)\in\mathbb{R}^{\mathrm{2}} ,{a}<{b} \\ $$$${f}\left({z}−{a}\right)={f}\left({b}−{z}\right)\mathrm{sin}\:\frac{{z}\pi}{{b}−{a}} \\ $$$${f}\left({z}\right)={z}^{\mathrm{2}} ,\Re\left({z}\right)\geqslant\frac{{a}+{b}}{\mathrm{2}} \\ $$$${f}\left({z}\right)=\mathrm{0},{z}=? \\ $$
Commented by Rasheed Soomro last updated on 20/Nov/15
$${Let}\:{a}=\mathrm{0}\:{then}\:{b}>\mathrm{0}: \\ $$$${f}\left({z}−{a}\right)={f}\left({b}−{z}\right)\mathrm{sin}\:\frac{{z}\pi}{{b}−{a}}\Rightarrow{f}\left({z}\right)={f}\left({b}−{z}\right)\mathrm{sin}\:\frac{{z}\pi}{{b}}=\mathrm{0}\:{if}\:{f}\left({z}\right)=\mathrm{0} \\ $$$$\Rightarrow{f}\left({b}−{z}\right)=\mathrm{0}\:\mid\left\{\:\mathrm{sin}\:\frac{{z}\pi}{{b}}=\mathrm{0}\Rightarrow\frac{{z}\pi}{{b}}={sin}^{−\mathrm{1}} \mathrm{0}\Rightarrow{z}=\mathrm{0}\right\} \\ $$$${f}\left({z}\right)=\mathrm{0}\Rightarrow{z}=\mathrm{0}\:{is}\:{one}\:{solution}. \\ $$$${For}\:{a}=\mathrm{0}\:{and}\:{b}>\mathrm{0} \\ $$$$\:\:\:\:\:{f}\left({z}\right)={f}\left({b}−{z}\right)\mathrm{sin}\:\frac{{z}\pi}{{b}}………….\left({i}\right) \\ $$$$\:\:\:\:\:{f}\left({b}−{z}\right)={f}\left({b}−\overline {{b}−{z}}\right){sin}\frac{\left({b}−{z}\right)\pi}{{b}}\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:={f}\left({z}\right){sin}\frac{\left({b}−{z}\right)\pi}{{b}}…….\left({ii}\right) \\ $$$${From}\:\left({i}\right)\:{and}\:\left({ii}\right) \\ $$$$\:\:\:\:{f}\left({z}\right)=\left({f}\left({z}\right){sin}\frac{\left({b}−{z}\right)\pi}{{b}}\right)\mathrm{sin}\:\frac{{z}\pi}{{b}} \\ $$$$\:\:\:\:\left({sin}\frac{\left({b}−{z}\right)\pi}{{b}}\right)\left(\mathrm{sin}\:\frac{{z}\pi}{{b}}\right)=\mathrm{1} \\ $$$$\:\:\:\:\frac{\mathrm{1}}{\mathrm{2}}\left\{{cos}\left(\frac{\left({b}−{z}\right)\pi}{{b}}−\frac{{z}\pi}{{b}}\right)−{cos}\left(\frac{\left({b}−{z}\right)\pi}{{b}}+\frac{{z}\pi}{{b}}\right)\right\}=\mathrm{1} \\ $$$$\:\:\:\:\frac{\mathrm{1}}{\mathrm{2}}\left\{{cos}\left(\frac{\left({b}−{z}−{z}\right)\pi}{{b}}\right)−{cos}\left(\frac{\left({b}−{z}+{z}\right)\pi}{{b}}\right)\right\}=\mathrm{1} \\ $$$$\:\:\:\:{cos}\left(\frac{\left({b}−\mathrm{2}{z}\right)\pi}{{b}}\right)−{cos}\left(\left(\pi\right)=\mathrm{2}\right. \\ $$$$\:\:\:\:{cos}\left(\frac{\left({b}−\mathrm{2}{z}\right)\pi}{{b}}\right)+\mathrm{1}=\mathrm{2} \\ $$$$\:\:\:\:{cos}\left(\frac{\left({b}−\mathrm{2}{z}\right)\pi}{{b}}\right)=\mathrm{1} \\ $$$$\:\:\:\frac{\left({b}−\mathrm{2}{z}\right)\pi}{{b}}={cos}^{−\mathrm{1}} \left(\mathrm{1}\right)=\mathrm{0} \\ $$$$\:\:\:{b}−\mathrm{2}{z}=\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:{z}=\frac{{b}}{\mathrm{2}} \\ $$$$\mathcal{C}{ontinue} \\ $$$$\:\:\:\:\: \\ $$
Commented by Rasheed Soomro last updated on 20/Nov/15
$${The}\:{condition}\:\Re\left({z}\right)\geqslant\frac{{a}+{b}}{\mathrm{2}}\:\:{is}\:{only}\:{for}\:{f}\left({z}\right)={z}^{\mathrm{2}} \:\:? \\ $$
Commented by Rasheed Soomro last updated on 21/Nov/15
$${When}\:{you}\:{write}\:{statement}\:{and}\:{condition}\:{on}\: \\ $$$${one}\:{line}\:{it}\:{should}\:{mean}\:{that}\:{the}\:{condition}\:{is} \\ $$$${only}\:{for}\:{that}\:{statement}.{For}\:{example}\:{the}\:{condition} \\ $$$$\Re\left({z}\right)\geqslant\frac{{a}+{b}}{\mathrm{2}}\:{is}\:{only}\:{in}\:{case}\:{f}\left({z}\right)={z}^{\mathrm{2}} \:: \\ $$$${f}\left({z}\right)={z}^{\mathrm{2}} ,\Re\left({z}\right)\geqslant\frac{{a}+{b}}{\mathrm{2}} \\ $$$${Similarly}\:{in}\:{the}\:{following}\:{you}\:{are}\:{asked}\:{for}\:{z}\:{when} \\ $$$${f}\left({z}\right)=\mathrm{0}.\:\:: \\ $$$${f}\left({z}\right)=\mathrm{0},{z}=? \\ $$$$\mathcal{A}{m}\:\mathcal{I}\:\mathcal{R}{ight}? \\ $$
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](https://www.tinkutara.com/question/Q2459.png)
$${f}:\mathbb{C}\rightarrow\mathbb{C},\left({a},{b}\right)\in\mathbb{R}^{\mathrm{2}} ,{a}<{b} \\ $$$${f}\left({z}−{a}\right)={f}\left({b}−{z}\right)\mathrm{sin}\:\frac{{z}\pi}{{b}−{a}} \\ $$$${f}\left({z}\right)={z}^{\mathrm{2}} ,\Re\left({z}\right)\geqslant\frac{{a}+{b}}{\mathrm{2}} \\ $$$${f}\left({z}\right)=\mathrm{0},{z}=? \\ $$$$−−−−−−−−−−−−−−−−−−−−− \\ $$$${f}\left({z}−{a}\right)={f}\left({b}−{z}\right)\mathrm{sin}\:\frac{{z}\pi}{{b}−{a}}\:{is}\:{true}\:{for}\:{any}\:{real}\:{a},{b}\:{with}\:{a}<{b} \\ $$$${Let}\:{a}=\mathrm{0},{then}\:{b}>\mathrm{0} \\ $$$${f}\left({z}−{a}\right)={f}\left({b}−{z}\right)\mathrm{sin}\:\frac{{z}\pi}{{b}−{a}} \\ $$$$\:\:\:\:\:\:\:\:\Rightarrow{f}\left({z}\right)={f}\left({b}−{z}\right)\mathrm{sin}\:\frac{{z}\pi}{{b}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\Rightarrow\:{f}\left({b}−{z}\right)={f}\left({b}−\overline {{b}−{z}}\right)\mathrm{sin}\:\frac{{z}\pi}{{b}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:={f}\left({z}\right)\mathrm{sin}\:\frac{{z}\pi}{{b}} \\ $$$${f}\left({z}\right)={f}\left({z}\right)\mathrm{sin}\:\frac{{z}\pi}{{b}}\mathrm{sin}\:\frac{{z}\pi}{{b}} \\ $$$${f}\left({z}\right)−{f}\left({z}\right)\mathrm{sin}^{\mathrm{2}} \:\frac{{z}\pi}{{b}}=\mathrm{0} \\ $$$${f}\left({z}\right)\left(\mathrm{1}−\mathrm{sin}^{\mathrm{2}} \:\frac{{z}\pi}{{b}}\right)=\mathrm{0} \\ $$$${f}\left({z}\right)\left(\mathrm{1}−\mathrm{sin}\frac{{z}\pi}{{b}}\right)\left(\mathrm{1}+\mathrm{sin}\frac{{z}\pi}{{b}}\right)=\mathrm{0} \\ $$$${f}\left({z}\right)=\mathrm{0}\:\:\:\mid\:\:\:\:\:\mathrm{sin}\frac{\pi{z}}{{b}}\:=\mathrm{1}\:\:\mid\:\:\:\mathrm{sin}\:\frac{\pi{z}}{{b}}=−\mathrm{1} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mid\:\:\frac{\pi{z}}{{b}}=\mathrm{sin}^{−\mathrm{1}} \left(\mathrm{1}\right)\:\:\mid\:\:\frac{\pi{z}}{{b}}=\mathrm{sin}^{−\mathrm{1}} \left(−\mathrm{1}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mid\:\frac{\pi{z}}{{b}}=\frac{\pi}{\mathrm{2}}\:\:\mid\:\:\frac{\pi{z}}{{b}}=\frac{\mathrm{3}\pi}{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mid\:{z}=\frac{{b}}{\mathrm{2}}\:\:\:\:\:\mid\:{z}=\frac{\mathrm{3}{b}}{\mathrm{2}}\:\:\:\:\:\:\:\:{for}\:{a}=\mathrm{0}\:,{b}>\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mid\:\:\Re\left({z}\right)\geqslant\:\frac{{a}+{b}}{\mathrm{2}}=\frac{\mathrm{0}+{b}}{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mid\:\:\frac{{b}}{\mathrm{2}}\geqslant\frac{{b}}{\mathrm{2}}\:\:\:\:\:\:{Condition}\:{satisfied}\:{for}\:{z}=\frac{{b}}{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mid\:\frac{\mathrm{3}{b}}{\mathrm{2}}\geqslant\frac{{b}}{\mathrm{2}}\:\:\:\:\:\:\:\:{as}\:{well}\:{as}\:\:\:\:{for}\:{z}=\frac{\mathrm{3}{b}}{\mathrm{2}} \\ $$$$−−−−−−−−−−−−− \\ $$$${Let}\:{b}=\mathrm{0}\:\left[{a}<\mathrm{0}\right] \\ $$$${f}\left({z}−{a}\right)={f}\left({b}−{z}\right)\mathrm{sin}\:\frac{{z}\pi}{{b}−{a}} \\ $$$$\Rightarrow{f}\left({z}−{a}\right)={f}\left(−{z}\right)\mathrm{sin}\:\frac{{z}\pi}{−{a}} \\ $$$$\Rightarrow{f}\left({z}−{a}\right)=−{f}\left(−{z}\right)\mathrm{sin}\:\frac{{z}\pi}{{a}} \\ $$$${f}\left(−{z}\right)\:=−\frac{{f}\left({z}−{a}\right)}{\mathrm{sin}\:\frac{{z}\pi}{{a}}} \\ $$$$\:\:\:\:\:\:\:\:{f}\left({z}−{a}\right)={f}\left(−\overline {{a}−{z}}\right)=−\frac{{f}\left({a}−{z}−{a}\right)}{\mathrm{sin}\:\frac{\left({a}−{z}\right)\pi}{{a}}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=−\frac{{f}\left(−{z}\right)}{\mathrm{sin}\:\frac{\left({a}−{z}\right)\pi}{{a}}} \\ $$$${f}\left(−{z}\right)\:=−\frac{−\frac{{f}\left(−{z}\right)}{\mathrm{sin}\:\frac{\left({a}−{z}\right)\pi}{{a}}}}{\mathrm{sin}\:\frac{{z}\pi}{{a}}} \\ $$$${f}\left(−{z}\right)=\frac{{f}\left(−{z}\right)}{\mathrm{sin}\:\frac{\left({a}−{z}\right)\pi}{{a}}.\mathrm{sin}\:\frac{{z}\pi}{{a}}} \\ $$$${f}\left(−{z}\right)−\frac{{f}\left(−{z}\right)}{\mathrm{sin}\:\frac{\left({a}−{z}\right)\pi}{{a}}.\mathrm{sin}\:\frac{{z}\pi}{{a}}}=\mathrm{0} \\ $$$${f}\left(−{z}\right)\left[\mathrm{1}−\frac{\mathrm{1}}{\mathrm{sin}\:\frac{\left({a}−{z}\right)\pi}{{a}}.\mathrm{sin}\:\frac{{z}\pi}{{a}}}\right]=\mathrm{0} \\ $$$${f}\left(−{z}\right)=\mathrm{0}\:\:\mid\:\:\frac{\mathrm{1}}{\mathrm{sin}\:\frac{\left({a}−{z}\right)\pi}{{a}}.\mathrm{sin}\:\frac{{z}\pi}{{a}}}=\mathrm{1} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mid\:\:\mathrm{sin}\:\frac{\left({a}−{z}\right)\pi}{{a}}.\mathrm{sin}\:\frac{{z}\pi}{{a}}=\mathrm{1} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mid\:\frac{\mathrm{1}}{\mathrm{2}}\left\{{cos}\left(\frac{\left({a}−{z}−{z}\right)\pi}{{a}}\right)−{cos}\left(\frac{\left({a}−{z}+{z}\right)\pi}{{a}}\right\}=\mathrm{1}\right. \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mid\:\frac{\mathrm{1}}{\mathrm{2}}\left\{{cos}\left(\frac{\left({a}−\mathrm{2}{z}\right)\pi}{{a}}\right)−{cos}\left(\frac{\left({a}\right)\pi}{{a}}\right\}=\mathrm{1}\right. \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mid\:\frac{\mathrm{1}}{\mathrm{2}}\left\{{cos}\left(\frac{\left({a}−\mathrm{2}{z}\right)\pi}{{a}}\right)−{cos}\left(\pi\right)\right\}=\mathrm{1} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mid\:{cos}\left(\frac{\left({a}−\mathrm{2}{z}\right)\pi}{{a}}\right)+\mathrm{1}=\mathrm{2} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mid\:{cos}\left(\frac{\left({a}−\mathrm{2}{z}\right)\pi}{{a}}\right)=\mathrm{1} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mid\:\:\:\frac{\left({a}−\mathrm{2}{z}\right)\pi}{{a}}={cos}^{−\mathrm{1}} \left(\mathrm{1}\right)=\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mid\:\:{a}−\mathrm{2}{z}=\mathrm{0}\Rightarrow{z}=\frac{{a}}{\mathrm{2}}\:\:\:\:\:\left[{Satisfy}\:{the}\:{condition}\:\Re\left({z}\right)\geqslant\frac{{a}+{b}}{\mathrm{2}}\right]\:\left[{b}=\mathrm{0}\right. \\ $$$$ \\ $$
Commented by Rasheed Soomro last updated on 21/Nov/15
$${This}\:{is}\:{not}\:{full}\:{answer}.{It}\:{is}\:{for}\:{a}=\mathrm{0}\:{or}\:{b}=\mathrm{0}\:{and}\:{also} \\ $$$${z}\:{is}\:{pure}\:{real}. \\ $$$${Needs}\:{verification}\:\:{and}\:\:{attestation}. \\ $$
Answered by Rasheed Soomro last updated on 22/Nov/15
$${f}:\mathbb{C}\rightarrow\mathbb{C},\left({a},{b}\right)\in\mathbb{R}^{\mathrm{2}} ,{a}<{b} \\ $$$${f}\left({z}−{a}\right)={f}\left({b}−{z}\right)\mathrm{sin}\:\frac{{z}\pi}{{b}−{a}} \\ $$$${f}\left({z}\right)={z}^{\mathrm{2}} ,\Re\left({z}\right)\geqslant\frac{{a}+{b}}{\mathrm{2}} \\ $$$${f}\left({z}\right)=\mathrm{0},{z}=? \\ $$$$−−−−−−−−−−−−−−−−−−−−− \\ $$$${Let}\:{z}={x}+{iy} \\ $$$${f}\left({z}−{a}\right)={f}\left({b}−{z}\right)\mathrm{sin}\:\frac{{z}\pi}{{b}−{a}}\Rightarrow{f}\left(\left({x}+{iy}\right)−{a}\right)={f}\left({b}−\left({x}+{iy}\right)\right)\mathrm{sin}\:\frac{\left({x}+{iy}\right)\pi}{{b}−{a}} \\ $$$${Let}\:\:{b}−{a}=\mathrm{1}\:{then}\:{b}={a}+\mathrm{1} \\ $$$$\:\:{f}\left(\left({x}+{iy}\right)−{a}\right)={f}\left(\:{a}+\mathrm{1}−\left({x}+{iy}\right)\:\right)\mathrm{sin}\:\left({x}+{iy}\right) \\ $$$${Continue} \\ $$