How do you multiply e^(( pi )/ 8 i) * e^( 3 pi/2 i ) eπ8ie3π2i in trigonometric form?

1 Answer
Mar 6, 2017

0.383 - 0.924i0.3830.924i

Explanation:

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+isinxeix=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)eπ8ie3π2i=eπ8i+3π2i=e13π8i

Now, put this into Euler's formula,

e^((13pi)/8i) = cos((13pi)/8)+isin((13pi)/8)e13π8i=cos(13π8)+isin(13π8)

= 0.383 - 0.924i=0.3830.924i

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)=eπ8ie3π2i=

(cos(pi/8)+isin(pi/8))*(cos((3pi)/2)+isin((3pi)/2))(cos(π8)+isin(π8))(cos(3π2)+isin(3π2))

= cos(pi/8)cos((3pi)/2) + isin(pi/8)cos((3pi)/2) + isin((3pi)/2)cos(pi/8) - sin(pi/8)sin((3pi)/2)=cos(π8)cos(3π2)+isin(π8)cos(3π2)+isin(3π2)cos(π8)sin(π8)sin(3π2)

This is, obviously, pretty ugly.

We know from trig formulas that

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

and

cosAcosB-sinAsinB = cos(A+B)cosAcosBsinAsinB=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(π8+3π2)+isin(π8+3π2)

=cos((13pi)/8)+isin((13pi)/8) = 0.383 - 0.924i=cos(13π8)+isin(13π8)=0.3830.924i

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.