Bounded linear functional on a $C^*$ algebra I want to prove the following statement:

Let $\mathcal{A}$ be a unital $C^*$ algebra with unit $e_\mathcal{A}$ and let $\varphi \in \mathcal{A}^*$ be a bounded linear functional in $\mathcal{A}$ such that 
  $$||\varphi||_{\mathcal{A}^*} = \varphi(e_\mathcal{A})=1.$$
  Show that $\varphi$ is a state.

So all I have to show is that $\varphi$ is positiv, that means $\varphi(A^*A)\geq 0$ for all $A\in\mathcal{A}.$
I got the hint to show first that $\varphi(A)\in\mathbb{R}$ for all self-adjoint $A=A^*\in \mathcal{A}$ by considering $\varphi(A+i\gamma e_\mathcal{A})$ with $\gamma\in\mathbb{R},$ but I don't know how to show that and how can I apply this for the statement above?
 A: Suppose $A$ is positive, then
$${\big\|}\,\|A\|e_\mathcal{A}-A\,\big\|≤\|A\|,$$
which you can see by considering the spectrum of $A$. Specifically $\sigma(A)$ consists only of numbers $≥0$ and we have $\|A\|=\sup_{\lambda\in\sigma(A)}|\lambda|$ for normal elements.
If you assume $\varphi(A)<0$ then $\varphi(\|A\|e_\mathcal{A}-A)=\|A\|-\varphi(A)>\|A\|$, contradicting $\|\varphi\|=1$.
So you see that for any positive $A$, $\varphi(A)$ must also be positive or admit a non-vanishing imaginary component. But positive elements are self-adjoint and here your hint tells you that $\varphi(A)$ can never have non-vanishing imaginary component if $A$ is positive.
To see how the hint can be proven, suppose $A$ is self-adjoint with $\varphi(A)= x+ iy$ with $y\neq0$. Note that for real $\lambda$
$$\|A+i\lambda\, e_{\mathcal A}\|= \sqrt{\|A\|^2+\lambda^2}$$
and that 
$$|\varphi(A+i\lambda\, e_\mathcal{A})|=\sqrt{x^2+(\lambda+y)^2}.$$
It follows that
$$\frac{\varphi(A+i\lambda\,e_\mathcal{A})}{\|A+i\lambda\,e_\mathcal{A}\|}=\sqrt{\frac{x^2+(\lambda+y)^2}{\|A\|^2+\lambda^2}}.$$
If you do some analysis you will find that for very large $\lambda$ this is roughly $\left|\mathrm{sign}(\lambda)+\frac y\lambda\right|$, once again contradicting $\|\varphi\|=1$.
