How to show ${\dfrac{1}{2}}(e^{\lambda t}x + e^{\overline{\lambda}t}{\overline{x}})=Re(e^{\lambda t}x)$?

Question

I'm learning Linear Algebra with Applications 8th Edition by Steve J Leon.

On P.300, in Subject "Complex Eigenvalues", Section 6.2 "Systems of Linear Differential Equations", Chapter 6 "Eigenvalues", it says $" ... Y_1={\dfrac{1}{2}}(e^{\lambda t}x + e^{\overline{\lambda}t}{\overline{x}})=Re(e^{\lambda t}x) ... "$.

But how to show ${\dfrac{1}{2}}(e^{\lambda t}x + e^{\overline{\lambda}t}{\overline{x}})=Re(e^{\lambda t}x)$? ${\lambda},x\in\mathbb{C}, t\in\mathbb{R}$.

Answer 1

I will assume that $t$ is real. $e^{\overline {z}}=\overline {e^{z}}$. You can prove this either from the series expansion of $e^{z}$ or from the formula $e^{a+ib}=e^{a}(\cos\, b+i \sin \,b)$ for $a,b$ real. Apply this formula to $z=\lambda t$ and use the fact that $\zeta+\overline {\zeta}=2 Re(\zeta)$.

Answer 2

Hint: $$\frac12(z+\overline{z})=\Re{(z)}\qquad\forall z\in\mathbb{C}$$ $$\overline{e^z}=e^{\overline{z}}\qquad\forall z\in\mathbb{C}$$ $$\overline{a\cdot b}=\overline{a}\cdot\overline{b}\qquad\forall a,b\in\mathbb{C}$$