|
Welcome Guest | Login | |
|
Index
| Recent Threads
| Who's Online
| Register
| Search
| Help
| 
|
 |
WebGraphing.com Forum » List all forums » Forum: Precalculus and Trigonometry Homework Help » Thread: complex numbers |
|
Total posts in this thread: 2 |
[Add To My Favorites] [Watch this Thread] [Post new Thread] |
| Author |
|
chelsarm
|
z1-3sqrt2-3isqrt2 z2=-1+sqrt3i find: z1*z2, z1/z2, (z2)^6, the 5th roots of z1 |
||
|
|
pskinner
Joined: Apr 2, 2005 Posts: 797 Status: Offline |
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... ---------------------------------------- Principal Skinner |
||||
|
| [Show Thread Printable Version] [Post new Thread] |
Re: complex numbers