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

The complex number has Cartesian form

.

Step-by-step explanation:

First, we need to recall the definition of when is a complex number:

.

Then,

. (I)

Now, recall the definition of the complex exponential:

.

So,

(we use that .

Thus,

Now we group conveniently in the above expression:

.

Now, substituting this equality in (I) we get

.

Thus,

.

For 15 to 20, change the equation from y=mx+b into mx-y=-b
Example: 15) y=-5x+2 into -2=-5x-y or 2=5x+y
For 21-29, the numbers in the parenthesis are coordinates (x, y) so uh the equation y=mx+b plug in the m that is given and x and y and then solve for b to complete the equation
Ex: 21) (-8,3) m=2
Y=mx+b
3=2(3)+b
3=6+b
-3=b
So y=2x-3
Do the same to the others