Question #89196

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Answer 1 · 2016-11-24T00:57:08

#z_1= 2e^((3pi)/2i)#
#z_2 =8e^((4pi)/3i)#
#z_1z_2 = 16e^((5pi)/6i)#
#z_1/z_2=1/4e^(pi/6i)#

Using Euler's formula #e^(itheta) = cos(theta)+isin(theta)#.

#z_1 = -2i#

#= 2(-i)#
#= 2(cos((3pi)/2)+isin((3pi)/2))#
#= 2e^((3pi)/2i)#

#z_2 = -4-4sqrt(3)i#

#= 8(-1/2-sqrt(3)/2i)#
#= 8(cos((4pi)/3)+isin((4pi)/3))#
#=8e^((4pi)/3i)#

Next, using that #e^xe^y = e^(xy)# and #e^x/e^y = e^(x-y)#, we have

#z_1z_2 = 2e^((3pi)/2i)*8e^((4pi)/3i) = 16e^((3/2+4/3)pii)=16e^((17pi)/6i) = 16e^((5pi)/6i)#

where the last step uses #e^(itheta) = e^(i(theta+2pin)), n in ZZ#

and

#z_1/z_2 = (2e^((3pi)/2i))/(8e^((4pi)/3i)) = 1/4e^((3/2-4/3)pii)=1/4e^(pi/6i)#