Question

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

Answer 1

The complex number
`e^(\cos(2+31)) = \exp(\cos(2+3i))` has Cartesian form

`\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)`.

Explanation:

First, we need to recall the definition of
`\cos z` when
`z` is a complex number:

`\cos z = \cos(x+iy) = (e^(iz)+e^(-iz))/(2)`.

Then,

`\cos(2+3i) = (e^(i(2+31)) + e^(-i(2+31)))/(2) = (e^(2i-3)+e^(-2i+3))/(2)`. (I)

Now, recall the definition of the complex exponential:

`e^(z)=e^(x+iy) = e^x(\cos y +i\sin y)`.

So,

`e^(2i-3) = e^(-3)(\cos 2+i\sin 2)`

`e^(-2i+3) = e^(3)(\cos 2-i\sin 2)` (we use that
`\sin(-y)=-\sin y)`.

Thus,

`e^(2i-3)+e^(-2i+3) = e^(-3)\cos 2+ie^(-3)\sin 2 + e^(3)\cos 2-ie^(3)\sin 2)`

Now we group conveniently in the above expression:

`e^(2i-3)+e^(-2i+3) = (e^(-3)+e^(3))\cos 2 + i(e^(-3)-e^(3))\sin 2`.

Now, substituting this equality in (I) we get

`\cos(2+3i) = (e^(-3)+e^(3))/(2)\cos 2 -i(e^(3)-e^(-3))/(2)\sin 2 = \cosh 3\cos 2-i\sinh 3\sin 2`.

Thus,

`\exp\left(\cos(2+3i)\right) = \exp\left(\cosh 3\cos 2-i\sinh 3\sin 2\right)`

`\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]`.