Prove normal matrix is 4th power of some self-adjoint matrix

Let $A\in\mathbb{M}_n(\mathbb{C})$ be a matrix of order $n\geq 2$. Let $p_A(x) = (x-\lambda_1)\cdots (x-\lambda_n)$ be the characteristic polynomial of $A$ such that all $\lambda_i$ are positive real numbers.

Suppose that $A$ is a normal matrix. Prove that one can write $A=G^4$ for some self-adjoint matrix $G$.

My thinking is that since $A$ is normal, by Principal Axis Theorem, we have $A=UDU^\ast$, where $U$ is a unitary matrix and $D$ a diagonal matrix. However, I fail to see how to proceed. Could anyone help me out?

You have that $D$ has diagonal $\lambda_1,\ldots,\lambda_n$. Since the eigenvalues are assume positive, you have $D=G_0^4$, where $G_0$ is diagonal with diagonal entries $\lambda_1^{1/4},\ldots,\lambda_n^{1/4}$. Now take $G=UG_0U^*$.

You're on the right track. Find $H$ such that $H^4=D$ and use $H$ to define $G$.