Need help with deriving the eigen-value and eigen-func for a S-L PDE (1dim Heat EQ)

1. Nov 1, 2011

Powertravel

Hi,
I am struggling with the heat equation
ut = kuxx
with the boundary conditions
u(0,t) = u'(L,t) = 0
and initial condition
u(x,0) = f(x)
0 ≤ x ≤ L
0 ≤ t

I want to derive it's eigenvalue using complex analysis.

After separating the variables into u(x,t) = X(x)T(t) = XT and getting
T' + λkT = 0 (1)
X'' + λX = 0 (2)

It is easily shown that λ ≠ 0 because it only yields elementary solutions.

Equation (2)s Auxiliary Equation (AE) is
r = ± i*√(λ)

So
X = Acosh(rx) + Bsinh(rx)
and
X' = Asinh(rx) + Bcosh(rx)

Using the Boundrary Conditions I get
X(0)=0 $\Rightarrow$ A = 0 and B ≠ 0
X'(L) = 0 $\Rightarrow$ cosh(rx) = 0
so
X = Bsinh(rx)

Now comes my first question:

My textbook says that cosh(rx) only is zero when
λ = (((2n -1)$\pi$)/(2*L))2 n = 1,2,3,... (3)
Can't it also be
λ = (((1 + 2n)$\pi$)/(2*L))2 n = 0,1,2,... ? (4)
Why is it as (3) instead of (4) and will (4) cause problems if I want to expand u(x,t) in a sine series?

Now for my second and primary question.

Using (3) I get
i*λL = n*$\pi$ - 0.5$\pi$ $\Rightarrow$
√(λ) = ((2n -1)$\pi$) / (2Li) = -i((2n -1)$\pi$)/(2L) $\Rightarrow$
λ = (-((2n -1)$\pi$)/(2L))2
That means λ < 0 so √(-λ) is real.
How can I get the eigen function of λ equal Csin(√(-λ)x) when X = sinh(√(-λ)x) ?