f-z-e-1-z-f-1-z-pi-z-sin-piz-f-z-z-2-z-1-2-f-z-0-z-

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

f (z) e^{1 - z} = f (1 - z) π^{z} \sin (π z)

f (z) = z^{2}, ℜ (z) ⩾ 1 / 2

f (z) = 0, z = ? ?

Commented by Yozzi last updated on 17/Nov/15

f (z) = 0 \Rightarrow z^{2} = 0

S u p p o s e z = a + b i; a, b \in R

\Rightarrow a^{2} - b^{2} + 2 i a b = 0

\Rightarrow a^{2} - b^{2} = 0 \Rightarrow a = \pm b a n d a b = 0 \Rightarrow a = 0 \lor b = 0

\Rightarrow a = b = 0 \Rightarrow R (z) = 0

B u t, R (z) ⩾ 1 / 2 \neq 0

∴ ∄ z \in C w i t h R (z) ⩾ 1 / 2 f o r f (z) = 0

T h i s i s t h e c a s e i f o n e l o o k s s o l e l y a t f (z) = 0

g i v e n t h a t f (z) = z^{2} .

Commented by Yozzi last updated on 17/Nov/15

I f (z) = 0, l o o k i n g a t t h e e q u a t i o n g i v e n,

\Rightarrow f (1 - z) π^{z} s i n (π z) = 0

\Rightarrow f (1 - z) = 0 \lor π^{z} = 0 \lor s i n (π z) = 0

π^{z} = e^{l n π^{z}} = e^{z l n π} = e^{(a + i b) l n π} = e^{a l n π} e^{i b l n π}

a, π, e \in R \Rightarrow e^{a l n π} \neq 0

∴ π^{z} = 0 \Rightarrow e^{i b l n π} = 0

∴ c o s (b l n π) + i s i n (b l n π) = 0

\Rightarrow c o s (b l n π) = 0 \land s i n (b l n π) = 0

∴ b l n π = \frac{(2 n + 1) π}{2}, b l n π = n π n \in Z

b = \frac{2 n + 1}{2 l n π} π, b = \frac{n π}{l n π}

I f \frac{2 n + 1}{2} = n \Rightarrow 2 n = 2 n + 1 \Rightarrow 0 = 1 (C o n t r a d i c t i o n)

∴ ∄ b \in R s u c h t h a t π^{z} = 0 \Rightarrow π^{z} \neq 0 \forall z \in C .

I f s i n (π z) = 0 \Rightarrow π z = n π \Rightarrow z = n; n \in Z

∴ a + b i = n \Rightarrow b = 0 \Rightarrow a = n .

∵ R (z) ⩾ 1 / 2 \Rightarrow a ⩾ 1 ∴ z = a w i t h a \in Z^{+} .

I f f (z) = z^{2} \Rightarrow f (1 - z) = {(1 - z)}^{2}

∴ f (1 - z) = 0 \Rightarrow {(1 - z)}^{2} = 0 \Rightarrow z = 1

Commented by RasheedAhmad last updated on 17/Nov/15

E x c e l l e n t!

Commented by RasheedAhmad last updated on 17/Nov/15

f (z) e^{1 - z} = f (1 - z) π^{z} \sin (π z)

z^{2} e^{1 - z} = f (1 - z) π^{z} \sin (π z)

I f z = 1, t h e e q u a t i o n {i s n}^{'} t s a t i s f i e d!

1 = 0 ?

Commented by Yozzi last updated on 17/Nov/15

Y u p, c e r t a i n l y . {I t}^{'} s f a l s e f o r z = a, a \in Z^{+} t o o .

G i v e n t h e c o n d i t i o n R (z) ⩾ 1 / 2

f o r f (z) = 0, t h e r e a r e n o s o l u t i o n s .

P e r h a p s o n e s h o u l d l o o k a t e^{1 - z} = 0 .

Commented by Yozzi last updated on 17/Nov/15

e^{1 - z} = e \times e^{- z} = e \times e^{- a - b i} = e^{1 - a} \times e^{- b i}

I f e^{1 - z} = 0 \Rightarrow e^{1 - a} e^{- b i} = 0 .

a, e \in R \Rightarrow e^{1 - a} \neq 0 \Rightarrow e^{- b i} = 0

\Rightarrow c o s b - i s i n b = 0

\Rightarrow c o t b - i = 0

c o t b = i w h i c h i s f a l s e i f b \in R .

\Rightarrow e^{1 - z} \neq 0 \forall z \in C .

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

f (z) e^{1 - z} = f (1 - z) π^{z} \sin (π z)

z = 0 \Rightarrow ℜ (z) ⪈ 1 / 2

f (z) e = f (1) \times π \times 0 = 1 \times π \times 0 = 0 \Rightarrow f (z) = 0

So z = 0 is a valid solution .

Commented by Yozzi last updated on 17/Nov/15

Y e s .

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

z = 0 ⇎ ℜ (z) ⪈ 1 / 2

f (0) e^{1 - 0} = f (1) \cdot π \cdot \sin (0)

f (0) \times e = 1^{2} \times π \times 0 = 0

General solution : z = - n π, n \in N \cup {0}

z = - n π \Rightarrow ℜ (z) ⪈ 1 / 2 \Rightarrow f (z) = 0