Find e^cos(2+3i) as a complex number expressed in Cartesian form.

Answer:The complex number [tex]e^{\cos(2+31)} = \exp(\cos(2+3i))[/tex] has Cartesian form[tex]\exp\left(\cosh 3\cos 2\right)\cos(\sinh 3\sin 2)-i\exp\left(\cosh 3\cos 2\right)\sin(\sinh 3\sin 2)[/tex].Step-by-step explanation:First, we need to recall the definition of [tex]\cos z[/tex] when [tex]z[/tex] is a complex number:[tex]\cos z = \cos(x+iy) = \frac{e^{iz}+e^{-iz}}{2}[/tex].Then,[tex]\cos(2+3i) = \frac{e^{i(2+31)} + e^{-i(2+31)}}{2} = \frac{e^{2i-3}+e^{-2i+3}}{2}[/tex]. (I)Now, recall the definition of the complex exponential:[tex]e^{z}=e^{x+iy} = e^x(\cos y +i\sin y)[/tex].So,[tex]e^{2i-3} = e^{-3}(\cos 2+i\sin 2)[/tex][tex]e^{-2i+3} = e^{3}(\cos 2-i\sin 2)[/tex] (we use that [tex]\sin(-y)=-\sin y)[/tex].Thus, [tex]e^{2i-3}+e^{-2i+3} = e^{-3}\cos 2+ie^{-3}\sin 2 + e^{3}\cos 2-ie^{3}\sin 2)[/tex]Now we group conveniently in the above expression:[tex]e^{2i-3}+e^{-2i+3} = (e^{-3}+e^{3})\cos 2 + i(e^{-3}-e^{3})\sin 2[/tex].Now, substituting this equality in (I) we get[tex]\cos(2+3i) = \frac{e^{-3}+e^{3}}{2}\cos 2 -i\frac{e^{3}-e^{-3}}{2}\sin 2 = \cosh 3\cos 2-i\sinh 3\sin 2[/tex].Thus,[tex]\exp\left(\cos(2+3i)\right) = \exp\left(\cosh 3\cos 2-i\sinh 3\sin 2\right)[/tex][tex]\exp\left(\cos(2+3i)\right) = \exp\left(\cosh 3\cos 2\right)\left[ \cos(\sinh 3\sin 2)-i\sin(\sinh 3\sin 2)\right][/tex].