Menu Close

find-zeros-function-f-z-1-cosz-




Question Number 145822 by Mrsof last updated on 08/Jul/21
find zeros function f(z)=1−cosz
$${find}\:{zeros}\:{function}\:{f}\left({z}\right)=\mathrm{1}−{cosz} \\ $$
Commented by Mrsof last updated on 08/Jul/21
???????
$$??????? \\ $$
Answered by mathmax by abdo last updated on 09/Jul/21
z_k =2kπ  are roots  f(z)=0 ⇒1−cosz=0 ⇒cosz=1 ⇒((e^(iz)  +e^(−iz) )/2)=1 ⇒e^(iz) +e^(−iz) −2=0  ⇒u+u^(−1) −2=0    (u=e^(iz) ) ⇒u^2  +1−2u=0 ⇒(u−1)^2 =0 ⇒u=1  e^(iz) =1 =e^(i(2kπ))  ⇒z_k =2kπ  (k∈Z)are roots of f
$$\mathrm{z}_{\mathrm{k}} =\mathrm{2k}\pi\:\:\mathrm{are}\:\mathrm{roots} \\ $$$$\mathrm{f}\left(\mathrm{z}\right)=\mathrm{0}\:\Rightarrow\mathrm{1}−\mathrm{cosz}=\mathrm{0}\:\Rightarrow\mathrm{cosz}=\mathrm{1}\:\Rightarrow\frac{\mathrm{e}^{\mathrm{iz}} \:+\mathrm{e}^{−\mathrm{iz}} }{\mathrm{2}}=\mathrm{1}\:\Rightarrow\mathrm{e}^{\mathrm{iz}} +\mathrm{e}^{−\mathrm{iz}} −\mathrm{2}=\mathrm{0} \\ $$$$\Rightarrow\mathrm{u}+\mathrm{u}^{−\mathrm{1}} −\mathrm{2}=\mathrm{0}\:\:\:\:\left(\mathrm{u}=\mathrm{e}^{\mathrm{iz}} \right)\:\Rightarrow\mathrm{u}^{\mathrm{2}} \:+\mathrm{1}−\mathrm{2u}=\mathrm{0}\:\Rightarrow\left(\mathrm{u}−\mathrm{1}\right)^{\mathrm{2}} =\mathrm{0}\:\Rightarrow\mathrm{u}=\mathrm{1} \\ $$$$\mathrm{e}^{\mathrm{iz}} =\mathrm{1}\:=\mathrm{e}^{\mathrm{i}\left(\mathrm{2k}\pi\right)} \:\Rightarrow\mathrm{z}_{\mathrm{k}} =\mathrm{2k}\pi\:\:\left(\mathrm{k}\in\mathrm{Z}\right)\mathrm{are}\:\mathrm{roots}\:\mathrm{of}\:\mathrm{f} \\ $$

Leave a Reply

Your email address will not be published. Required fields are marked *