Matrix Exponential and Conjugation of Jordan Normal Form and Real Normal Form

Question

Given a diagonalizable matrix $A \in \mathbb{R^{N \times N}}$, we can decompose $A$ in following terms:

1. $A = V \Lambda_{C} V^{-1}$ where $V, \Lambda_{C} \in \mathbb{C^{N \times N}}$
2. $A = Q \Lambda_{R} Q^{-1}$ where $Q, \Lambda_{R} \in \mathbb{R^{N \times N}}$

In a matrix form, we can see the following relationship in a simplist case:

$ \Lambda_{C} = \begin{pmatrix} a + bi & 0 \\ 0 & a - bi \end{pmatrix}, \quad \Lambda_{R} = \begin{pmatrix} a & b \\ -b & a \end{pmatrix}, \quad P \Lambda_{R} P^{-1} = \Lambda_{C} \quad (P = V^{-1}Q) $

Now, the question arises when we take a matrix exponential.

$exp(\Lambda_C) = \begin{pmatrix} e^{a + bi} & 0 \\ 0 & e^{a - bi} \end{pmatrix} = \begin{pmatrix} e^a (cos(b) + isin(b)) & 0 \\ 0 & e^a (cos(b) - isin(b)) \end{pmatrix} $

Again, we can change the basis of this matrix to make it real form. Let's say we have another matrix $K \in \mathbb{C}$:

$ Kexp(\Lambda_C)K^{-1} = \begin{pmatrix} e^a cos(b) & e^a sin(b) \\ -e^a sin(b) & e^a cos(b)) \end{pmatrix} = \Lambda $

Here, I am not sure on following questions:

1. Is $exp(\Lambda_{R}) = \Lambda$?
2. Related question. Is $K = P = V^{-1}Q$?

Core concept of the question is finding a conjugate matrix that changes Jordan Normal Form to Real Normal Form.