erfc with complex number

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?

Answer 1

[Matt Fig]

A quick search of the FEX turned up these two. There may be others...

File 1

File 2

[Dilip Jose]

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]

[Dilip Jose]

i have tried something but couldn't....pls find the attachemnt

[Dilip Jose]

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]

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]

yes sir now im using fadf.m for erfc...but now symbolic variable is exist inside fadf ...error shwing wat to do

[Walter Roberson]

https://www.mathworks.com/matlabcentral/fileexchange/47801-the-voigt-complex-error-function-second-version for fadf?

[Dilip Jose]

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]

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]

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]

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]

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]

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]

thank you so much sir ...got some results...hope the best.