# why we find discontinuity when we plot this analytic function \begin{align*} f(z) =(0.5+000i)+(0.5000 + 0.8660i) z+(-0.2500+0.4330i)z^2 \end{align*}

조회 수: 1 (최근 30일)
Aisha Mohamed 2022년 6월 18일
댓글: Aisha Mohamed 2022년 6월 19일
I used the code which is provided by TORSTEN(Thank you so much) to plot this function:
$f(z) =(0.5+000i)+(0.5000 + 0.8660i) z+(-0.2500+0.4330i)z^2$ .I write this function in polar cordinate as follows
f = @(x,y) 1*( 0.5.*exp(1i*0)+((x+1i*y).^1*exp(1i*pi/3))+(0.5)*(x+1i*y).^2*exp(1i*-pi/3)) .But the figer is discontinuty as yo can see :
I used this code
clc
close all
clear all
f = @(x,y) (0.5)+((x+1i*y).*exp(1i*pi/3))+(0.5)*(x+1i*y).^2*exp(-1i*pi/3);
r = 0.2:0.01:0.9;
phi = 0:0.01:2*pi;
[R,PHI] = meshgrid(r,phi);
X = R.*cos(PHI);
Y = R.*sin(PHI);
F=f(X,Y);
figure
subplot(2,2,1)
surf(X,Y,abs(f(X,Y)),'EdgeColor','none')
view(2)
colorbar
title('|f(z)|')
xlabel('Z_R')
ylabel('Z_I')
% Zlabel('|f(z)|')
grid on
subplot(2,2,2)
surf(X,Y,angle(f(X,Y)),'EdgeColor','none')
view(2)
colorbar
title('phase of f(z)')
xlabel('Z_R')
ylabel('Z_I')
% Zlabel('|f(z)|')
grid on
subplot(2,2,3)
surf(X,Y,angle(f(X,Y)),'EdgeColor','none')
view(2)
colorbar
title('phase of f(z),view(2)')
xlabel('Z_R')
ylabel('Z_I')
grid on
I think there is some mistake inmy writing of the function f, because this code work perfictly with other function like
f = @(x,y) 0.5./(x+1i*y).^3.*((x+1i*y).^3-exp(1i*pi/3))./((x+1i*y)-exp(-1i*pi/3)).
I appriciate any help
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Torsten 2022년 6월 18일
It's you who made the "hole" by choosing
r = 0.2:0.01:0.9;
r = 0:0.01:0.9;
Aisha Mohamed 2022년 6월 19일
Thank you so much Torsten and Sam

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