erfc with complex number
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Hi all
I want to compute erfc(.) of a complex number. How can I do that? Is there any way to use erfc(.) of matlab?
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Matt Fig
2011-5-10
2 个投票
14 个评论
Dilip Jose
2020-1-30
sir will you pls tel me how to define erfc(.)....in terms of complex number ....im having my rsults interms exponentiam and complementary error function ....dont know how to plot this.....struggling for 3 months 
Dilip Jose
2020-1-30
Dilip Jose
2020-1-30
i have tried something but couldn't....pls find the attachemnt
Dilip Jose
2020-1-30
close all
syms z;
gr=5;
gc=5;
M=1;
w=0.5;
pr=7;
sc=2.01;
t=0.4;
m=(M+(2*1i*w));
s1=z;
s2=(2*sqrt(t));
x=((s1)./(s2));
a1=(x*t);
a2=(2*m*(sqrt(pi)));
a=((m)./(pr-1));
b=((m)./(sc-1));
x=((s1)./(s2));
u1=((x^2+(m*t))*t);
u2=4*m;
d1=(x*(sqrt(t)))*(1-(4*m*t));
d2=(8*m^3/2);
v1=gr;
v2=(a*(1-pr));
k1=gc;
k2=(b*(1-sc));
h1=exp(a*t);
h2=2;
e1=(exp(2*x*(sqrt(m*t))));
e2=(erfc(x+(sqrt(m*t))));
e3=(exp(-2*x*(sqrt(m*t))));
e4=(erfc(x-(sqrt(m*t))));
b1=(exp(2*x*((sqrt((m+a)*t)))));
b2=(erfc(x+(sqrt((m+a)*t))));
b3=(exp(-2*x*(sqrt((m+a)*t))));
b4=(erfc(x-(sqrt((m+a)*t))));
b5=(exp(2*x*((sqrt((m+b)*t)))));
b6=(erfc(x+(sqrt((m+b)*t))));
b7=(exp(-2*x*(sqrt((m+b)*t))));
b8=(erfc(x-(sqrt((m+b)*t))));
e5=exp(-(x^2+(m*t)));
c1=(exp(2*x*(sqrt(pr*a*t))));
c2=(erfc(x*(sqrt(pr)+(sqrt(a*t)))));
c3=((exp(-2*x*(sqrt(pr*a*t)))));
c4=(erfc(x*(sqrt(pr)-(sqrt(a*t)))));
l1=(exp(2*x*(sqrt(sc*b*t))));
l2=(erfc(x*(sqrt(sc)+(sqrt(b*t)))));
l3=((exp(-2*x*(sqrt(sc*b*t)))));
l4=(erfc(x*(sqrt(sc)-(sqrt(b*t)))));
f1=(exp(b*t));
f2=2;
l=((a1)./(a2));
f=((f1)./(f2));
h=((h1)./(h2));
j1=((u1)./(u2));
j2=((d1)./(d2));
j3=((v1)./(v2));
j4=((k1)./(k2));
q1=(2*(j1*((e1*e2)+(e3*e4))+(j2*((e3*e4)-(e1*e2))-(l*e5))));
q2=((j3+j4)*(1/2)*((e1*e2)+(e3*e4)));
q3=((j3*(h*(b1*b2)+(b3*b4))));
q4=((j4*(f*(b5*b6)+(b7*b8))));
q5=(j3*(erfc(x*sqrt(pr))));
q6=(j4*erfc(x*sqrt(sc)));
q7=(j3*(h*((c1*c2)+(c3*c4))));
q8=(j4*(f*((l1*l2)+(l3*l4))));
q=(q1+q2-q3-q4-q5-q6+q7+q8);
fplot(z,q);
xlim([0 5])
ylim([0 1])
hold on
title('Axial velocity profiles for different values of gr & gc');
hold off
Walter Roberson
2020-1-30
The symbolic toolbox can take erfc of a complex number as long as it is passed in symbolic form. erfc(sym(.3-.2i)) for example.
Dilip Jose
2020-2-5
yes sir now im using fadf.m for erfc...but now symbolic variable is exist inside fadf ...error shwing wat to do
Dilip Jose
2020-2-5
编辑：Dilip Jose
2020-2-5
sir i have installed package.....in that erfc change to fadf.....in fadf(symbolic variable)...what to do....
syms z
dt=0.01;t1=0;t2=5;t=(t1:dt:t2);
u=0.2;
w=0.5;
M=1;
m=((M+(2*1i*w))*u);
k=(fadf(z./(2*(u).^.5)+(M+(2*1i*w))*u).^.5);
re_k=real(k);im_k=imag(k);
plot(re_z,im_k);
Walter Roberson
2020-2-5
fadf is not designed for symbolic use.
If you have the symbolic toolbox then it already handles erfc of a complex number as long as the number is passed in symbolic form.
Dilip Jose
2020-2-5
yes sir thank you...in my result z is infinite plane ....by defining sym z...i have done my plotting....but in velocity profile equation ...erfc of complex number exist....fadf is used to define erfc of complex...getting error...how to define z infinite plane extent..need a support.
Walter Roberson
2020-2-5
In your original code at https://www.mathworks.com/matlabcentral/answers/7153-erfc-with-complex-number#comment_791526 you have
fplot(z,q);
where z is a plain symbolic variable and q is a symbolic expression in z. However, q is complex valued and fplot() cannot plot complex valued things. You can, though, do things like
fplot3(z, real(q), imag(q))
You cannot use fplot to plot over the entire z plane -- for one thing the imaginary component of q goes to about - 9.878*10^6730 by the time z is -1e4 .
Walter Roberson
2020-2-5
fafd needs to be able to test
ind_neg = imag(z)<0; % if some imag(z) values are negative, then ...
z(ind_neg) = conj(z(ind_neg)); % ... bring them to the upper-half plane
FF = zeros(size(z)); % define array
ind_ext = abs(z)>8; % external indices
ind_band = ~ind_ext & imag(z)<5*10^-3; % narrow band indices
It cannot test those things for symbolic z, because z could have any value to be determined later.
To work with fafd, you will need to create specific numeric z and calculate based upon those.
syms z
dt=0.01;t1=0;t2=5;t=(t1:dt:t2);
u=0.2;
w=0.5;
M=1;
m=((M+(2*1i*w))*u);
k=(fadf(z./(2*(u).^.5)+(M+(2*1i*w))*u).^.5);
re_k=real(k);im_k=imag(k);
plot(re_z,im_k);
You do not use t after you create it, so perhaps your z should be set to be t ?
Walter Roberson
2020-2-5
To check, you have
d2=(8*m^3/2);
Is it possible that that should be
d2=(8*m^(3/2));
The current expression would be equivalent to the clearer
d2=4*m^3;
Or even 8/2 instead of 4 if that somehow made documentation sense. With the /2 where it is and with the leading multiplier already even, people are going to wonder.
Dilip Jose
2020-2-13
thank you so much sir ...got some results...hope the best.
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