Hyperbolic version Euler formula

We know that $e^{ix}=\cos x+i\sin x$$e^{ix}=\cos x+i\sin x$

And we can find a hyperbolic version of that.

Consider the hyperbolic complex number $\mathbb{H}\cong {\displaystyle \frac{\mathbb{R}[x]}{(x^2-1)}}$$\mathbb H\cong \displaystyle \frac{\mathbb R[x]}{(x^2-1)}$

$(a_0+b_{0}x)(a_1+b_{1}x)=a_0{a}_1+(a_0{b}_1+a_1{b}_0)x+b_0{b}_1{x}^2$$(a_0+b_0 x)(a_1+b_1x)=a_0a_1+(a_0b_1+a_1b_0)x+b_0b_1x^2$

$=a_0{a}_1-b_0{b}_1+(a_0{b}_1+a_1{b}_0)x+b_0{b}_1{x}^2-b_0{b}_1$$=a_0a_1-b_0b_1+(a_0b_1+a_1b_0)x+b_0b_1 x^2-b_0b_1$

$=a_0{a}_1-b_0{b}_1+(a_0{b}_1+a_1{b}_0)x+b_0{b}_1(x^2-1)=a_0{a}_1-b_0{b}_1+(a_0{b}_1+a_1{b}_0)x$$=a_0a_1-b_0b_1+(a_0b_1+a_1b_0)x+b_0b_1(x^2-1)=a_0a_1-b_0b_1+(a_0b_1+a_1b_0)x$

So we can represent the hyperbolic complex number as $x+yj$$x+yj$, $j^2=1$$j^2=1$

And then consider $f(x)=\cosh (x)+j\sinh (x)$$f(x)=\cosh(x)+j\sinh(x)$, we can find ${\displaystyle \frac{df}{dx}=jf}$$\displaystyle\frac{df}{dx}=jf$

Thus we get $e^{jx}=\cosh (x)+j\sinh (x)$$e^{jx}=\cosh(x)+j\sinh(x)$

You can check it by considering the Taylor Series.