Show that bilinear form is coercive

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I have got the following problem. Let $G$ be bounded and connected. For $u,v \in H^1(G)$ and $\lambda \in \mathbb{C}$ define $B(u,v) := \langle \nabla u, \nabla v \rangle$ and $B_{\lambda}(u,v) := \langle \nabla u, \nabla v \rangle - \lambda \langle u , v \rangle$ where $\langle \cdot , \cdot \rangle$ is the $L^2(G)$ dot product. Define $D(A) := \lbrace u \in H^1(G) \vert \exists f \in L^2(G) : \forall v \in H^1(G):B(v,u) = \langle v,f \rangle \rbrace$.

I showed already that the set $D(A)$ implies the Neumann boundary conditions $\frac{\partial u}{\partial \vec{n}} = 0$ for the Poisson Equation $\triangle u = f$.

and I showed

For $u \in D(A)$ exists exactly one $f_u \in L^2(G)$ such that $\forall v \in H^1(G) : B(u,v) = \langle v,f_u \rangle$.

Now I'm stuck at the following

I need to show that

  1. $\forall \lambda \in \mathbb{C}$ with $Re(\lambda) < 0$: $B_{\lambda}$ is coercive.
  2. For all $0 \neq \lambda \in i \mathbb{R}$: $\vert B_{\lambda} (u,u) \vert \geq \alpha \Vert u \Vert_{H^1}$ for $\alpha > 0$.

We defined coercive in this context as $Re(B(u,u)) \geq p \Vert u \Vert_{H^1}^2$ for $p > 0$.

Any tip would be great.