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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:C→C,(a,b)∈R2,a<bf(z−a)=f(b−z)sinzπb−af(z)=z2,ℜ(z)⩾a+b2f(z)=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
Leta=0thenb>0:f(z−a)=f(b−z)sinzπb−a⇒f(z)=f(b−z)sinzπb=0iff(z)=0⇒f(b−z)=0∣{sinzπb=0⇒zπb=sin−10⇒z=0}f(z)=0⇒z=0isonesolution.Fora=0andb>0f(z)=f(b−z)sinzπb………….(i)f(b−z)=f(b−b−z―)sin(b−z)πb=f(z)sin(b−z)πb…….(ii)From(i)and(ii)f(z)=(f(z)sin(b−z)πb)sinzπb(sin(b−z)πb)(sinzπb)=112{cos((b−z)πb−zπb)−cos((b−z)πb+zπb)}=112{cos((b−z−z)πb)−cos((b−z+z)πb)}=1cos((b−2z)πb)−cos((π)=2cos((b−2z)πb)+1=2cos((b−2z)πb)=1(b−2z)πb=cos−1(1)=0b−2z=0z=b2Continue
Commented by Rasheed Soomro last updated on 20/Nov/15
The condition ℜ(z)≥((a+b)/2) is only for f(z)=z^2 ?
Theconditionℜ(z)⩾a+b2isonlyforf(z)=z2?
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?
Whenyouwritestatementandconditionononelineitshouldmeanthattheconditionisonlyforthatstatement.Forexampletheconditionℜ(z)⩾a+b2isonlyincasef(z)=z2:f(z)=z2,ℜ(z)⩾a+b2Similarlyinthefollowingyouareaskedforzwhenf(z)=0.:f(z)=0,z=?AmIRight?
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:C→C,(a,b)∈R2,a<bf(z−a)=f(b−z)sinzπb−af(z)=z2,ℜ(z)⩾a+b2f(z)=0,z=?−−−−−−−−−−−−−−−−−−−−−f(z−a)=f(b−z)sinzπb−aistrueforanyreala,bwitha<bLeta=0,thenb>0f(z−a)=f(b−z)sinzπb−a⇒f(z)=f(b−z)sinzπb⇒f(b−z)=f(b−b−z―)sinzπb=f(z)sinzπbf(z)=f(z)sinzπbsinzπbf(z)−f(z)sin2zπb=0f(z)(1−sin2zπb)=0f(z)(1−sinzπb)(1+sinzπb)=0f(z)=0∣sinπzb=1∣sinπzb=−1∣πzb=sin−1(1)∣πzb=sin−1(−1)∣πzb=π2∣πzb=3π2∣z=b2∣z=3b2fora=0,b>0∣ℜ(z)⩾a+b2=0+b2∣b2⩾b2Conditionsatisfiedforz=b2∣3b2⩾b2aswellasforz=3b2−−−−−−−−−−−−−Letb=0[a<0]f(z−a)=f(b−z)sinzπb−a⇒f(z−a)=f(−z)sinzπ−a⇒f(z−a)=−f(−z)sinzπaf(−z)=−f(z−a)sinzπaf(z−a)=f(−a−z―)=−f(a−z−a)sin(a−z)πa=−f(−z)sin(a−z)πaf(−z)=−−f(−z)sin(a−z)πasinzπaf(−z)=f(−z)sin(a−z)πa.sinzπaf(−z)−f(−z)sin(a−z)πa.sinzπa=0f(−z)[1−1sin(a−z)πa.sinzπa]=0f(−z)=0∣1sin(a−z)πa.sinzπa=1∣sin(a−z)πa.sinzπa=1∣12{cos((a−z−z)πa)−cos((a−z+z)πa}=1∣12{cos((a−2z)πa)−cos((a)πa}=1∣12{cos((a−2z)πa)−cos(π)}=1∣cos((a−2z)πa)+1=2∣cos((a−2z)πa)=1∣(a−2z)πa=cos−1(1)=0∣a−2z=0⇒z=a2[Satisfytheconditionℜ(z)⩾a+b2][b=0
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.
Thisisnotfullanswer.Itisfora=0orb=0andalsozispurereal.Needsverificationandattestation.
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:C→C,(a,b)∈R2,a<bf(z−a)=f(b−z)sinzπb−af(z)=z2,ℜ(z)⩾a+b2f(z)=0,z=?−−−−−−−−−−−−−−−−−−−−−Letz=x+iyf(z−a)=f(b−z)sinzπb−a⇒f((x+iy)−a)=f(b−(x+iy))sin(x+iy)πb−aLetb−a=1thenb=a+1f((x+iy)−a)=f(a+1−(x+iy))sin(x+iy)Continue

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