MAT334-2018F > Quiz-7

Q7 TUT 0202

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Victor Ivrii:
Analysis on straight segments is not complete:

$x$ from $0$ to $R$; indeed, $f(x)$ stays real, but argument of real negative is not $2n\pi$, it is $(2n+1)\pi$. You need to check if the sign of $f(x)$ changes here.

$yi$ from $Ri$ to $0$: there could be an error in the multiple of $2n\pi$. In this case you see that $f(yi)$ is imaginary and does change sign. So one can say: "may be arg changes from $\pi/2$ to $-pi/2$ or may be to $3pi/2$, how do we know?" So you need to look at $f(yi+\varepsilon)$ with $0<\varepsilon\ll 1$ and look how its real part changes signs (or if it changes at all). Use $f(yi+\varepsilon)=f(yi)+ f'(yi) \varepsilon$

Heng Kan:
Please see the new attached scanned picture.   For yi on Ri to 0, as long as y is positive, f(yi) always lies in the fourth quardarnt.When R tends to be infinity, Re(f(iR)) = 7 and Im(f(iR)) tends to be negtive infinity. When R=0, f(iR) = 7. So f(iy) approximately rotates from negative imaginary axis to positive real axis counter-clockwisely.

Victor Ivrii:
Now it is a flawless analysis