Help with finding zeros of a complex function
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Hi there.
I have been given a task, where i need to find the zeros of the following complex function:

I have tried using fzero, but can't really make it work. I already know the answer to the task, which is that there are zeros in:
and in 
Do you guys know a way to make Matlab calculate this?
Thanks a lot!
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回答 (2 件)
Star Strider
2018 年 12 月 22 日
If you want reasonably close results for the roots:
g = @(z) z.*exp(z)./((z-pi).*cos(z));
zv = linspace(-10, 10, 5000)*pi;
gz = g(zv);
zci = @(v) find(v(:).*circshift(v(:), [-1 0]) <= 0); % Returns Approximate Zero-Crossing Indices Of Argument Vector
idxv = zci(gz);
for k1 = 1:numel(idxv)
b = [zv(idxv(k1)-1) 1; zv(idxv(k1)) 1] \ [gz(idxv(k1)-1); gz(idxv(k1))]; % Linear Fit Near Zero-Crossing
rt(k1) = -b(2)/b(1); % Interpolate To Precise Value
end
figure
plot(zv, g(zv))
hold on
plot(rt, zeros(size(rt)), 'xr')
hold off
grid
ylim([-5, 5])
fprintf('\nAs multiples of ‘pi’:\n')
fprintf('\t%8.2f\n', rt/pi)
fprintf('\nAs multiples of ‘n’:\n')
fprintf('\t%8.2f\n', (rt-pi/2)/pi)
Using fzero, the loop becomes:
for k1 = 1:numel(idxv)
rt(k1) = fzero(g, [zv(idxv(k1)-1),zv(idxv(k1)+1)]); % Use ‘fzero’ To Interpolate To Precise Value
end
with much more precise results.
It is likely not possible to do this without a loop, simply because it is necessary to iterate over all the zero-crossings.
3 件のコメント
madhan ravi
2018 年 12 月 22 日
編集済み: madhan ravi
2018 年 12 月 22 日
Use fsolve(), vpasolve() or solve() and ezplot() or fplot() to view where the zero occurs..
2 件のコメント
madhan ravi
2018 年 12 月 22 日
Hi. Thanks for the answer.
I just tried that - but am getting the wrong result. I have inserted a picture of what i have done:

madhan ravi
2018 年 12 月 22 日
First thing that you need to know is uploading a picture is never useful.. upload your code
Second thing what do you observe from the graph?
Third thing why do you say the result is wrong?
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