Plotting a function with different range of values for arguments

I am trying to plot a function with two different inputs, so its two plots in one graph actually. There are some constants which I have predefined. The function I want to plot is

$f(k,|T|^2)(=|T|^2/|R|^2)=|(e^{ik}-e^{-ik})/(1+(\delta-e^{ik})(e^{ik}-\nu)|^2$

The plots to be produced are for k=pi/2 and k=-pi/2, so I predefined these things in the code. However, when I try to plot the function for T, I get error saying that Undefined function or variable 'T'. The code I wrote is:

clear all
  v0=-2.5;
  epsilon=0.05;
  v1 = v0*(1+epsilon);
  v2 = v0*(1-epsilon);
  g = 1;

    k=pi/2
    l=-k
    w=-2*cos(k)
    w1=-2*cos(l)
  nu=v2-w+g.*T.^2;
  delta=v1-w+g.*T.^2.*(1-2.*nu*cos(k)+nu.^2);
  nu1=v1-w1+g.*T.^2;
  delta1=v2-w1+g.*T.^2.*(1-2.*nu*cos(k)+nu.^2);

    fplot(@(T) abs((exp(i*k)-exp(-i*k))/(1+(nu-exp(i*k)).*(exp(i*k)-delta)))^2,[0,5],'b')
    hold on
    fplot(@(T) abs((exp(i*l)-exp(-i*l))/(1+(nu1-exp(i*l)).*(exp(i*l)-delta1)))^2,[0,5],'b')
    hold off

So whats going wrong here? Secondly, is it the correct way for plotting functions which have an injected argument?

UPDATE

Ok this has been fixed by an ugly way with the following code: (Is there a better neater way to to do it?)

clear all
v0=-2.5;
epsilon=0.05;
v1 = v0*(1+epsilon);
v2 = v0*(1-epsilon);
g = 1;
T=[0:0.01:5];
k=pi/2
l=-k
w=-2*cos(k)
w1=-2*cos(l)
nu=v2-w+g.*T.^2;
delta=v1-w+g.*T.^2.*(1-2.*nu*cos(k)+nu.^2);
nu1=v1-w1+g.*T.^2;
delta1=v2-w1+g.*T.^2.*(1-2.*nu*cos(k)+nu.^2);
ftn1=abs((exp(i*k)-exp(-i*k))./(1+(nu-exp(i*k)).*(exp(i*k)-delta))).^2;
ftn2=abs((exp(i*l)-exp(-i*l))./(1+(nu1-exp(i*l)).*(exp(i*l)-delta1))).^2;
plot(T,ftn1,T,ftn2)

Answer 1

[Walter Roberson]

 nu = @(T) v2-w+g.*T.^2;
 delta = @(T) v1-w+g.*T.^2.*(1-2.*nu(T)*cos(k)+nu(T).^2); 
  nu1 = @(T) v1-w1+g.*T.^2;
  delta1 = @(T) v2-w1+g.*T.^2.*(1-2.*nu(T)*cos(k)+nu(T).^2);

 fplot(@(T) abs((exp(i*k)-exp(-i*k))/(1+(nu(T)-exp(i*k)).*(exp(i*k)-delta(T))))^2,[0,5],'b')
 hold on
 fplot(@(T) abs((exp(i*l)-exp(-i*l))/(1+(nu1(T)-exp(i*l)).*(exp(i*l)-delta1(T))))^2,[0,5],'b')
 hold off

[AtoZ]

@Walter Roberson, Thanks.. It works :-)