Question and Answers Forum

All Questions      Topic List

Relation and Functions Questions

Previous in All Question      Next in All Question      

Previous in Relation and Functions      Next in Relation and Functions      

Question Number 2344 by 123456 last updated on 17/Nov/15

f(z)e^(1−z) =f(1−z)π^z sin (πz)  f(z)=z^2 ,ℜ(z)≥1/2  f(z)=0,z=??

$${f}\left({z}\right){e}^{\mathrm{1}−{z}} ={f}\left(\mathrm{1}−{z}\right)\pi^{{z}} \mathrm{sin}\:\left(\pi{z}\right) \\ $$$${f}\left({z}\right)={z}^{\mathrm{2}} ,\Re\left({z}\right)\geqslant\mathrm{1}/\mathrm{2} \\ $$$${f}\left({z}\right)=\mathrm{0},{z}=?? \\ $$

Commented by Yozzi last updated on 17/Nov/15

f(z)=0⇒z^2 =0  Suppose z=a+bi ; a,b∈R  ⇒a^2 −b^2 +2iab=0  ⇒a^2 −b^2 =0⇒a=±b and ab=0⇒a=0∨b=0  ⇒ a=b=0 ⇒R(z)=0  But, R(z)≥1/2≠0  ∴ ∄z∈C with R(z)≥1/2 for f(z)=0  This is the case if one looks solely at f(z)=0  given that f(z)=z^2 .

$${f}\left({z}\right)=\mathrm{0}\Rightarrow{z}^{\mathrm{2}} =\mathrm{0} \\ $$$${Suppose}\:{z}={a}+{bi}\:;\:{a},{b}\in\mathbb{R} \\ $$$$\Rightarrow{a}^{\mathrm{2}} −{b}^{\mathrm{2}} +\mathrm{2}{iab}=\mathrm{0} \\ $$$$\Rightarrow{a}^{\mathrm{2}} −{b}^{\mathrm{2}} =\mathrm{0}\Rightarrow{a}=\pm{b}\:{and}\:{ab}=\mathrm{0}\Rightarrow{a}=\mathrm{0}\vee{b}=\mathrm{0} \\ $$$$\Rightarrow\:{a}={b}=\mathrm{0}\:\Rightarrow\mathfrak{R}\left({z}\right)=\mathrm{0} \\ $$$${But},\:\mathfrak{R}\left({z}\right)\geqslant\mathrm{1}/\mathrm{2}\neq\mathrm{0} \\ $$$$\therefore\:\nexists{z}\in\mathbb{C}\:{with}\:\mathfrak{R}\left({z}\right)\geqslant\mathrm{1}/\mathrm{2}\:{for}\:{f}\left({z}\right)=\mathrm{0} \\ $$$${This}\:{is}\:{the}\:{case}\:{if}\:{one}\:{looks}\:{solely}\:{at}\:{f}\left({z}\right)=\mathrm{0} \\ $$$${given}\:{that}\:{f}\left({z}\right)={z}^{\mathrm{2}} . \\ $$

Commented by Yozzi last updated on 17/Nov/15

If (z)=0, looking at the equation given,  ⇒f(1−z)π^z sin(πz)=0  ⇒f(1−z)=0 ∨ π^z =0 ∨ sin(πz)=0  π^z =e^(lnπ^z ) =e^(zlnπ) =e^((a+ib)lnπ) =e^(alnπ) e^(iblnπ)   a,π,e∈R⇒e^(alnπ) ≠0  ∴π^z =0⇒e^(iblnπ) =0  ∴cos(blnπ)+isin(blnπ)=0  ⇒cos(blnπ)=0 ∧ sin(blnπ)=0  ∴ blnπ=(((2n+1)π)/2),   blnπ=nπ  n∈Z  b=((2n+1)/(2lnπ))π   ,  b=((nπ)/(lnπ))  If  ((2n+1)/2)=n⇒2n=2n+1⇒0=1 (Contradiction)  ∴ ∄b∈R such that π^z =0⇒π^z ≠0 ∀z∈C.    If sin(πz)=0⇒ πz=nπ⇒z=n; n∈Z  ∴ a+bi=n⇒b=0⇒a=n.  ∵ R(z)≥1/2⇒ a≥1 ∴ z=a with a∈Z^+ .    If f(z)=z^2  ⇒f(1−z)=(1−z)^2   ∴ f(1−z)=0⇒(1−z)^2 =0⇒z=1

$${If}\:\left({z}\right)=\mathrm{0},\:{looking}\:{at}\:{the}\:{equation}\:{given}, \\ $$$$\Rightarrow{f}\left(\mathrm{1}−{z}\right)\pi^{{z}} {sin}\left(\pi{z}\right)=\mathrm{0} \\ $$$$\Rightarrow{f}\left(\mathrm{1}−{z}\right)=\mathrm{0}\:\vee\:\pi^{{z}} =\mathrm{0}\:\vee\:{sin}\left(\pi{z}\right)=\mathrm{0} \\ $$$$\pi^{{z}} ={e}^{{ln}\pi^{{z}} } ={e}^{{zln}\pi} ={e}^{\left({a}+{ib}\right){ln}\pi} ={e}^{{aln}\pi} {e}^{{ibln}\pi} \\ $$$${a},\pi,{e}\in\mathbb{R}\Rightarrow{e}^{{aln}\pi} \neq\mathrm{0} \\ $$$$\therefore\pi^{{z}} =\mathrm{0}\Rightarrow{e}^{{ibln}\pi} =\mathrm{0} \\ $$$$\therefore{cos}\left({bln}\pi\right)+{isin}\left({bln}\pi\right)=\mathrm{0} \\ $$$$\Rightarrow{cos}\left({bln}\pi\right)=\mathrm{0}\:\wedge\:{sin}\left({bln}\pi\right)=\mathrm{0} \\ $$$$\therefore\:{bln}\pi=\frac{\left(\mathrm{2}{n}+\mathrm{1}\right)\pi}{\mathrm{2}},\:\:\:{bln}\pi={n}\pi\:\:{n}\in\mathbb{Z} \\ $$$${b}=\frac{\mathrm{2}{n}+\mathrm{1}}{\mathrm{2}{ln}\pi}\pi\:\:\:,\:\:{b}=\frac{{n}\pi}{{ln}\pi} \\ $$$${If}\:\:\frac{\mathrm{2}{n}+\mathrm{1}}{\mathrm{2}}={n}\Rightarrow\mathrm{2}{n}=\mathrm{2}{n}+\mathrm{1}\Rightarrow\mathrm{0}=\mathrm{1}\:\left({Contradiction}\right) \\ $$$$\therefore\:\nexists{b}\in\mathbb{R}\:{such}\:{that}\:\pi^{{z}} =\mathrm{0}\Rightarrow\pi^{{z}} \neq\mathrm{0}\:\forall{z}\in\mathbb{C}. \\ $$$$ \\ $$$${If}\:{sin}\left(\pi{z}\right)=\mathrm{0}\Rightarrow\:\pi{z}={n}\pi\Rightarrow{z}={n};\:{n}\in\mathbb{Z} \\ $$$$\therefore\:{a}+{bi}={n}\Rightarrow{b}=\mathrm{0}\Rightarrow{a}={n}. \\ $$$$\because\:\mathfrak{R}\left({z}\right)\geqslant\mathrm{1}/\mathrm{2}\Rightarrow\:{a}\geqslant\mathrm{1}\:\therefore\:{z}={a}\:{with}\:{a}\in\mathbb{Z}^{+} . \\ $$$$ \\ $$$${If}\:{f}\left({z}\right)={z}^{\mathrm{2}} \:\Rightarrow{f}\left(\mathrm{1}−{z}\right)=\left(\mathrm{1}−{z}\right)^{\mathrm{2}} \\ $$$$\therefore\:{f}\left(\mathrm{1}−{z}\right)=\mathrm{0}\Rightarrow\left(\mathrm{1}−{z}\right)^{\mathrm{2}} =\mathrm{0}\Rightarrow{z}=\mathrm{1} \\ $$

Commented by RasheedAhmad last updated on 17/Nov/15

Excellent!

$$\mathcal{E}{xcellent}! \\ $$

Commented by RasheedAhmad last updated on 17/Nov/15

f(z)e^(1−z) =f(1−z)π^z sin (πz)  z^2 e^(1−z) =f(1−z)π^z sin (πz)  If z=1,the equation isn′t satisfied!           1=0?

$${f}\left({z}\right){e}^{\mathrm{1}−{z}} ={f}\left(\mathrm{1}−{z}\right)\pi^{{z}} \mathrm{sin}\:\left(\pi{z}\right) \\ $$$${z}^{\mathrm{2}} {e}^{\mathrm{1}−{z}} ={f}\left(\mathrm{1}−{z}\right)\pi^{{z}} \mathrm{sin}\:\left(\pi{z}\right) \\ $$$${If}\:{z}=\mathrm{1},{the}\:{equation}\:{isn}'{t}\:{satisfied}! \\ $$$$\:\:\:\:\:\:\:\:\:\mathrm{1}=\mathrm{0}? \\ $$

Commented by Yozzi last updated on 17/Nov/15

Yup, certainly. It′s false for z=a, a∈Z^+  too.  Given the condition R(z)≥1/2  for f(z)=0, there are no solutions.  Perhaps one should look at e^(1−z) =0.

$${Yup},\:{certainly}.\:{It}'{s}\:{false}\:{for}\:{z}={a},\:{a}\in\mathbb{Z}^{+} \:{too}. \\ $$$${Given}\:{the}\:{condition}\:\mathfrak{R}\left({z}\right)\geqslant\mathrm{1}/\mathrm{2} \\ $$$${for}\:{f}\left({z}\right)=\mathrm{0},\:{there}\:{are}\:{no}\:{solutions}. \\ $$$${Perhaps}\:{one}\:{should}\:{look}\:{at}\:{e}^{\mathrm{1}−{z}} =\mathrm{0}. \\ $$

Commented by Yozzi last updated on 17/Nov/15

e^(1−z) =e×e^(−z) =e×e^(−a−bi) =e^(1−a) ×e^(−bi)   If e^(1−z) =0⇒e^(1−a) e^(−bi) =0.  a,e∈R⇒e^(1−a) ≠0 ⇒e^(−bi) =0  ⇒cosb−isinb=0  ⇒cotb−i=0  cotb=i which is false if b∈R.  ⇒e^(1−z) ≠0 ∀z∈C.

$${e}^{\mathrm{1}−{z}} ={e}×{e}^{−{z}} ={e}×{e}^{−{a}−{bi}} ={e}^{\mathrm{1}−{a}} ×{e}^{−{bi}} \\ $$$${If}\:{e}^{\mathrm{1}−{z}} =\mathrm{0}\Rightarrow{e}^{\mathrm{1}−{a}} {e}^{−{bi}} =\mathrm{0}. \\ $$$${a},{e}\in\mathbb{R}\Rightarrow{e}^{\mathrm{1}−{a}} \neq\mathrm{0}\:\Rightarrow{e}^{−{bi}} =\mathrm{0} \\ $$$$\Rightarrow{cosb}−{isinb}=\mathrm{0} \\ $$$$\Rightarrow{cotb}−{i}=\mathrm{0} \\ $$$${cotb}={i}\:{which}\:{is}\:{false}\:{if}\:{b}\in\mathbb{R}. \\ $$$$\Rightarrow{e}^{\mathrm{1}−{z}} \neq\mathrm{0}\:\forall{z}\in\mathbb{C}. \\ $$

Commented by prakash jain last updated on 17/Nov/15

f(z)e^(1−z) =f(1−z)π^z sin (πz)  z=0⇒ℜ(z)≱1/2  f(z)e=f(1)×π×0=1×π×0=0⇒f(z)=0  So z=0 is a valid solution.

$${f}\left({z}\right){e}^{\mathrm{1}−{z}} ={f}\left(\mathrm{1}−{z}\right)\pi^{{z}} \mathrm{sin}\:\left(\pi{z}\right) \\ $$$${z}=\mathrm{0}\Rightarrow\Re\left({z}\right)\ngeqslant\mathrm{1}/\mathrm{2} \\ $$$${f}\left({z}\right){e}={f}\left(\mathrm{1}\right)×\pi×\mathrm{0}=\mathrm{1}×\pi×\mathrm{0}=\mathrm{0}\Rightarrow{f}\left({z}\right)=\mathrm{0} \\ $$$$\mathrm{So}\:{z}=\mathrm{0}\:\mathrm{is}\:\mathrm{a}\:\mathrm{valid}\:\mathrm{solution}. \\ $$

Commented by Yozzi last updated on 17/Nov/15

Yes.

$${Yes}. \\ $$

Answered by prakash jain last updated on 17/Nov/15

z=0⇎ℜ(z)≱1/2  f(0)e^(1−0) =f(1)∙π∙sin (0)  f(0)×e=1^2 ×π×0=0  General solution: z=−nπ, n∈N∪{0}  z=−nπ⇒ℜ(z)≱1/2⇒f(z)=0

$${z}=\mathrm{0}\nLeftrightarrow\Re\left({z}\right)\ngeqslant\mathrm{1}/\mathrm{2} \\ $$$${f}\left(\mathrm{0}\right){e}^{\mathrm{1}−\mathrm{0}} ={f}\left(\mathrm{1}\right)\centerdot\pi\centerdot\mathrm{sin}\:\left(\mathrm{0}\right) \\ $$$${f}\left(\mathrm{0}\right)×{e}=\mathrm{1}^{\mathrm{2}} ×\pi×\mathrm{0}=\mathrm{0} \\ $$$$\mathrm{General}\:\mathrm{solution}:\:{z}=−{n}\pi,\:{n}\in\mathbb{N}\cup\left\{\mathrm{0}\right\} \\ $$$${z}=−{n}\pi\Rightarrow\Re\left({z}\right)\ngeqslant\mathrm{1}/\mathrm{2}\Rightarrow{f}\left({z}\right)=\mathrm{0} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com