How do you multiply #e^(( pi )/ 8 i) * e^( 3 pi/2 i ) # in trigonometric form?

You can do this in two ways, one of which is shorter than the other.

For both you must know the formula

#e^(ix) = cosx+isinx#

which is known as Euler's formula.

I would multiply the two terms together first, and then put into trigonometric form, rather than putting them into trigonometric form and then multiplying the result, though either works.

Multiplying two power-terms means you just add the power, so

#e^(pi/8i)*e^((3pi)/2i) = e^(pi/8i+(3pi)/2i) = e^((13pi)/8i)#

Now, put this into Euler's formula,

#e^((13pi)/8i) = cos((13pi)/8)+isin((13pi)/8)#

# = 0.383 - 0.924i#

Conversely, you could put them into trig form first, and use trig addition formulas and such to simplify it.

#e^(pi/8i)*e^((3pi)/2i)=#

#(cos(pi/8)+isin(pi/8))*(cos((3pi)/2)+isin((3pi)/2))#

#= cos(pi/8)cos((3pi)/2) + isin(pi/8)cos((3pi)/2) + isin((3pi)/2)cos(pi/8) - sin(pi/8)sin((3pi)/2)#

This is, obviously, pretty ugly.

We know from trig formulas that

#sinAcosB+sinBcosA = sin(A+B)#

and

#cosAcosB-sinAsinB = cos(A+B)#

and, if you look closely, we have both of these patterns in our equation, so we actually have

#cos(pi/8 + (3pi)/2) + isin(pi/8 + (3pi)/2)#

#=cos((13pi)/8)+isin((13pi)/8) = 0.383 - 0.924i#

which is exactly what we had above.

As you can see, this method was much longer. Easier to just add up the indices and convert into trig functions. Nifty.