Author Topic: Q7 TUT 0202  (Read 10178 times)

Victor Ivrii

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Q7 TUT 0202
« on: November 30, 2018, 03:56:08 PM »
Using argument principle along line on the picture, calculate the number of zeroes of the following function in the first quadrant:
f(z)=z7+6z3+7.


Siying Li

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Re: Q7 TUT 0202
« Reply #1 on: November 30, 2018, 04:11:34 PM »
Since f(z)=z7+6z3+7

Then in the first quadrant,
When z goes from 0 to R on real axis,
z=x  f(x)=x7+6x3+7 f(0)=7,arg(f(z)) =0 f(R)= +  as R goes to+,arg(f(z)) =0

When z goes from 0 to iR on imaginary axis,
z=iy f(z)=(iy)7+6(iy)3+7=7i(y7+6y3)=7, arg(f(z)) =arc(tan(y7+6y37) )
 
When z is in between,
z=Reit, 0tπ2f(z)=(Reit)7+6(Reit)3+7=R7ei7t+6R3ei3t+7=R7(ei7t+6ei3tR4+7R4)arg(f(z)) 7twhen t=0, 7t=0 when t=π2, 7t=2π+32π
 
The net change of argument is  overall 4π, so 2 zeros in the first quadrant

sishan

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Re: Q7 TUT 0202
« Reply #2 on: December 01, 2018, 03:11:55 AM »
Let f(z) = u + iv = z7+6z3+7

Let z = Reiθ, and  0θπ2R

f(z) is analytic at all points except z = . Therefore, it is analytic within and upon the complementary of first quadrant.

when z = x,

f(z)=u+iv=x7+6x3+7

argf=tan1(vu)=tan1(0x7+6x3+7) = 0, x 0

Therefore, argf=0


when z = Reiθ, 0θπ2R
 
 f(z) = R7e7iθ(1+6R4e4iθ+7R7e7iθ)
 
 when R, fR7e7iθ  and arg f = 7θ
 
 argf=7(π20)=7π2


when z = iy,

f(z) = u + iv =7+6x3+7

argf=tan1(vu)=tan1(y76y37)=π2  from 0



argf=7π2+π2=4π

Thus, the angle change is 4π, and the number of zero in the first quadrant is 2.
« Last Edit: December 01, 2018, 03:20:53 AM by sishan »

Victor Ivrii

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Re: Q7 TUT 0202
« Reply #3 on: December 01, 2018, 01:33:26 PM »
everybody is either wrong or missing something

Heng Kan

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Re: Q7 TUT 0202
« Reply #4 on: December 01, 2018, 03:40:52 PM »
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.
« Last Edit: December 01, 2018, 05:44:36 PM by Heng Kan »

Victor Ivrii

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Re: Q7 TUT 0202
« Reply #5 on: December 01, 2018, 04:06:02 PM »
Analysis on straight segments is not complete:

x from 0 to R; indeed, f(x) stays real, but argument of real negative is not 2nπ, it is (2n+1)π. 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 2nπ. In this case you see that f(yi) is imaginary and does change sign. So one can say: "may be arg changes from π/2 to pi/2 or may be to 3pi/2, how do we know?" So you need to look at f(yi+ε) with 0<ε1 and look how its real part changes signs (or if it changes at all). Use f(yi+ε)=f(yi)+f(yi)ε

Heng Kan

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Re: Q7 TUT 0202
« Reply #6 on: December 01, 2018, 06:01:19 PM »
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

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Re: Q7 TUT 0202
« Reply #7 on: December 01, 2018, 06:32:03 PM »
Now it is a flawless analysis