Toronto Math Forum
MAT3342020F => MAT334Tests and Quizzes => Quiz 5 => Topic started by: Jiaqi Bi on November 05, 2020, 08:24:07 AM

$\textbf{Problem}$:
Give the order of each of the zeros of the given function:
$$e^{2z}3e^z4$$
$\textbf{Solution}$:
$f(z)=e^{2z}3e^z4=0$
Let $w=e^z$, we get $w^23w4=(w4)(w+1)$
Then resubstitute $e^z=w$
We get $(e^z4)(e^z+1)=0 \Rightarrow e^z=4\ \text{and}\ e^z=1$
Solve for $z$, we get $z=log(4)=ln(4)+i(2k\pi)\ \text{and}\ z=log(1)
=ln(1)+i(2k\pi)
=\pi i+i(2k\pi)
=i(\pi+2k\pi)$
To find orders: $f'(z)=2e^{2z}3e^z$
Substitute $e^z=4\ \text{and}\ 1$ into $f'(z)$
$2\times 163\times 4 = 20 \neq 0, 2+3=5\neq 0$
Since first derivative does not equal to $0$ for both $z$, we conclude order $= 1$
Hence, the order is $1$ for $z=ln(4)+i(2k\pi)\ \text{and}\ z=i(\pi+2k\pi)$