### Author Topic: Lec 0101- Quiz 7 C  (Read 1200 times)

#### Jiaqi Bi

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##### Lec 0101- Quiz 7 C
« on: December 17, 2020, 06:51:38 AM »
Question: Using argument principle along line on the picture, calculate the number of zeroes of the following function in the left half-plane:
$z+a=e^z, (a>0)$

(The graph will be attached.)
(Plus: I think Professor Victor originally wants to give us $a>1$ rather than $a>0$?)
Solution:
\begin{align} h(iy)&=iy+a-e^{iy}\\ \nonumber &=iy+a-cos(y)-isin(y)\\ \nonumber &=(a-cos(y))+i(y-sin(y)) \end{align}
If $a>0$, we cannot conclude anything for $\text{Re} h(iy)$, but if $a>1$, then $\text{Re} h(iy)$ is always positive because the range of $cos(y)$ is consistently from $-1$ to $1$.
$\text{Im}h(iy)$ will increase when $y$ goes from $-R$ to $R$.
\begin{align} h(\text{Re}^{it})=\text{Re}^{it}+a=e^{\text{Re}^{it}} \end{align}
Where t is from $\frac{\pi}{2}$ to $\frac{3\pi}{2}$, and $z$ goes from $iR$ to $-iR$. $h(z)$ in this circumstance has been travelled a counterclockwise circuit.
Therefore, the argument for $h(z)$ should be $2\pi$.
By The Argument Principle, $\frac{1}{2\pi}\cdot 2\pi=1$.
Hence, $h(z)$ has a total of one zero in this plane.
« Last Edit: December 19, 2020, 10:56:58 AM by Jiaqi Bi »

#### Jiaqi Bi

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##### Re: Lec 0101- Quiz 7 C
« Reply #1 on: December 17, 2020, 07:43:36 AM »
Question: Using argument principle along line on the picture, calculate the number of zeroes of the following function in the left half-plane:
$z+a=e^z, (a>0)$

(Plus: I think Professor Victor originally wants to give us $a>1$ rather than $a>0$?)
(The graph will be attached.)
Solution:
\begin{align} h(iy)&=iy+a-e^{iy}\\ &=iy+a-cos(y)-isin(y)\\ &=(a-cos(y))+i(y-sin(y)) \end{align}
If $a>0$, we cannot conclude anything for $\text{Re} h(iy)$, but if $a>1$, then $\text{Re} h(iy)$ is always positive because the range of $cos(y)$ is consistently from $-1$ to $1$.
$\text{Im}h(iy)$ will increase when $y$ goes from $-R$ to $R$.
\begin{align} h(\text{Re}^{it})=\text{Re}^{it}+a=e^{\text{Re}^{it}} \end{align}
Where t is from $\frac{\pi}{2}$ to $\frac{3\pi}{2}$, and $z$ goes from $iR$ to $-iR$. $h(z)$ in this circumstance has been travelled a counterclockwise circuit.
Therefore, the argument for $h(z)$ should be $2\pi$.
By The Argument Principle, $\frac{1}{2\pi}\cdot 2\pi=1$.
Hence, $h(z)$ has a total of one zero in this plane.