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chelsarm

USA
Joined: Dec 6, 2010
Posts: 2
Status: Offline

z1-3sqrt2-3isqrt2
z2=-1+sqrt3i
find: z1*z2, z1/z2, (z2)^6, the 5th roots of z1
 [Dec 6, 2010 7:54:36 PM] [Link]
pskinner

Joined: Apr 2, 2005
Posts: 797
Status: Offline

 z1-3sqrt2-3isqrt2z2=-1+sqrt3ifind: z1*z2, z1/z2, (z2)^6, the 5th roots of z1

I gather you mean z1=3*sqrt(2)-3*i*sqrt(2).

To find z1*z2, treat this as (a+bi)*(c+di):

z1*z2=(3*sqrt(2)-3*i*sqrt(2))*(-1+sqrt(3)*i)

=-3*sqrt(2)+3*i*sqrt(2)+3*i*sqrt(6)-3*(i^2)*sqrt(6)

=-3*sqrt(2)+3*i*sqrt(2)+3*i*sqrt(6)+3*sqrt(6)

=(-3*sqrt(2)+3*sqrt(6))+(3*sqrt(2)+3*sqrt(6))*i

To find z1/z2, multiply both numerator an denominator by the conjugate of the denominator, (-1-sqrt(3)*i) so you will get a real number in the denominator...

The third item requires use of DeMoivre's Theorem, which in turn requires you to rewrite z2 in terms of polar form:

z2=-1+sqrt(3)*i=r(cos(theta)+i*sin(theta))

where r=|z2|=sqrt((-1)^2+sqrt(3)^2)=sqrt(4)=2
so
2cos(theta)=-1, so theta=arccos(-1/2)=2pi/3

so

z2=2(cos(2pi/3)+i*sin(2pi/3)

the rest is a matter of plugging into appropriate equations...
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Principal Skinner
 [Dec 6, 2010 8:45:51 PM] [Link]