Asymptotic expansion of $\int_{-\infty}^{\infty}z^{1/2} e^{-z^2+2i\lambda z}dz$ as $\lambda \to \infty$ I was given the following problem.

Derive the leading term of an asymptotic expansion of the integral
$\displaystyle\int_{-\infty}^{\infty}z^{1/2} e^{-z^2+2i\lambda z}dz$ as $\lambda
 \to \infty$.

I felt like I should use the method of stationary phase or the method of steepest descent, but Since $2iz$ does not have a saddle point, it seems like it is not applicable.
My attempt
Suppose $f(z) =e^{-z^2+2i\lambda z}$.
Denote the integral by $I(\lambda)$.
Since $e^{-z^2}$ decays exponentially as $z\to \infty$, the contribution to the integral comes from the neighborhood of $z=0$. Since $e^{-z^2} \sim 1$ as $z\to 0$, we have $f(z) \sim z^{1/2}e^{2i\lambda z}$ as $z\to 0$. Thus,
$\displaystyle I(\lambda) \sim \int_{-\infty}^{\infty}z^{1/2}e^{2i\lambda z}dz$
Now, split the integral in two:
$\displaystyle \int_{-\infty}^{\infty}z^{1/2}e^{2i\lambda z}dz  =\int_{-\infty}^{0}z^{1/2}e^{2i\lambda z}dz + \int_{0}^{\infty}z^{1/2}e^{2i\lambda z}dz$
By direct calculation, we have
$\displaystyle \int_{-\infty}^{0}z^{1/2}e^{2i\lambda z}dz = \frac{i}{(2\lambda)^{3/2}}\Gamma\left(\frac{3}{2}\right)e^{-3\pi i/4}$
$\displaystyle \int_{0}^{\infty}z^{1/2}e^{2i\lambda z}dz = \frac{1}{(2\lambda)^{3/2}}\Gamma\left(\frac{3}{2}\right)e^{3\pi i/4}$
Since $ie^{-3\pi i/4}+e^{3\pi i/4}=0$, we have $I(\lambda) \sim 0$.

I think this is not correct since it is not possible to have $0$ as a leading term. However, I don't know what to do instead.
Any help, hint, or suggestion?
 A: I will assume that we are integrating above the branch cut, i.e., $\arg z=\pi$ on the portion of the contour leading to $0$. With the change of variables $z = \lambda t$, the integral becomes
$$
I(\lambda)=\lambda ^{3/2} \int_{ - \infty }^{ + \infty } {t^{1/2} e^{ - \lambda ^2 (t - 2i)t} dt} .
$$
We now push the contour upwards so that it goes through the saddle point at $t=i$. With $t=i+s$ ($s$ real) the integral becomes
$$
I(\lambda)=\lambda ^{3/2} e^{ - \lambda ^2 } \int_{ - \infty }^{ + \infty } {(i + s)^{1/2} e^{ - \lambda ^2 s^2 } ds} .
$$
Applying the saddle point method, the leading order asymptotics is found to be
$$
I(\lambda)\sim e^{\pi i/4} \sqrt {\pi \lambda }\, e^{ - \lambda ^2 }
$$
as $\lambda \to +\infty$.
A: Too long for comments.
After @Gary's answer, looking for an asymptotics, if $\lambda >0$,
$$I(\lambda)=\frac{i+1}{2 \sqrt{2}}e^{-\frac{\lambda ^2}{2}} \lambda ^{3/2}
   \left(K_{\frac{1}{4}}\left(\frac{\lambda
   ^2}{2}\right)+K_{\frac{3}{4}}\left(\frac{\lambda ^2}{2}\right)\right)$$
$$K_{\frac{1}{4}}\left(\frac{\lambda
   ^2}{2}\right)+K_{\frac{3}{4}}\left(\frac{\lambda ^2}{2}\right)=\frac{2 \sqrt{\pi } }{\lambda }e^{-\frac{\lambda ^2}{2}}\Bigg[1+\frac{1}{16 \lambda ^2}-\frac{15}{512 \lambda
   ^4}+O\left(\frac{1}{\lambda^6 }\right) \Bigg]$$
$$I(\lambda)=(1+i)\sqrt{\frac{\lambda\pi }{2}} e^{-\lambda ^2} \Bigg[1+\frac{1}{16 \lambda ^2}-\frac{15}{512 \lambda
   ^4}+O\left(\frac{1}{\lambda^6 }\right) \Bigg]$$
What is interesting too is that, if $\lambda <0$, we also have an explicit expression for $I(\lambda)$ involving only modified Bessel functions of the first kind.
