Q5 TUT 0101

[Victor Ivrii]

Give the order of each of the zeros of the given function:
$$
z^2(1-\cos(z)).
$$

[Tianfangtong Zhang]

Let $z^2(1-\cos(z)) = 0$

Then $z = 0$ or $\cos(z) = 1$

thus $z = 2k\pi$

 

case $1$: $z = 0$

let $f(z) = z^2$ and $h(z) = 1-\cos(z)$

Then $f(0)=0$

$f^{'}(0) = 2z|_{z=0} = 0$

$f^{''}(0)=2\neq 0$

thus order = 2

Then $h(0) = 0$

$h^{'}(0) = 2\sin(z)|_{z=0} = 0$

$f^{''}(0)=\cos(z)\neq 0$

thus order = 2

Therefore order(0) = 4



case $2$: $z = 2k\pi$ ($k\neq 0)$

let $f(z) = z^2$ and $h(z) = 1-\cos(z)$

Then $f(z) = z^2 = (2k\pi)^2 \neq 0$

thus order = 0

$h(z) = 1- \cos(z) = 0 $

$h^{'}(z) = \sin(z) = 0$

$h^{''} = \cos(z) = 0$

thus order = 2

Therefore order($2k\pi$) = 2, $k\neq0$

[Yuechen Huang]

z² (1 − cos(z))=0

z = 0 or cos (z) = 1

z = 0 or 𝑧 = 2𝑘𝜋

When z = 0

let 𝑓(𝑧) = z², ℎ(𝑧) = 1− cos (𝑧)

then 𝑓(0) = 0, 𝑓′(0) = 2𝑧 = 0, 𝑓″(0) = 2 ≠ 0

thus order = 2

ℎ(0)=0, ℎ′(0)= 2 sin(0)=0, h″(0)= cos(𝑧)≠0

thus order = 2

Therefore, order (0) = 4

When 𝑧 = 2𝑘𝜋 (𝑘≠0)

let 𝑓(𝑧) = z² and ℎ(𝑧) = 1−cos(𝑧)

𝑓(𝑧) = z² = (2𝑘𝜋)² ≠ 0
 
thus order = 0 

ℎ(𝑧)= 1− cos(𝑧)=0, ℎ′(𝑧)= sin(𝑧)=0, ℎ″= cos(𝑧)=0

thus order = 2

Therefore order (2𝑘𝜋) = 2, 𝑘 ≠ 0

[Victor Ivrii]

Yuechen
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