A community for students.
Here's the question you clicked on:
 0 viewing
EarthCitizen
 4 years ago
((3+j3)^3*(1j)^4)/((1+j(3)^1/2)^9)
EarthCitizen
 4 years ago
((3+j3)^3*(1j)^4)/((1+j(3)^1/2)^9)

This Question is Closed

EarthCitizen
 4 years ago
Best ResponseYou've already chosen the best response.0\[((3+j3)^{3} *(1j)^{4})/(1+j \sqrt{3})^{9}\]

asnaseer
 4 years ago
Best ResponseYou've already chosen the best response.1Just digesting that equation :)

asnaseer
 4 years ago
Best ResponseYou've already chosen the best response.1ok  it would be best to try and break it down a bit to simplify it first...

asnaseer
 4 years ago
Best ResponseYou've already chosen the best response.1\[(3+3j)^3=(3(1+j))^3=3^3(1+j)^3=27(1+j)^3\]

asnaseer
 4 years ago
Best ResponseYou've already chosen the best response.1next simplification...

asnaseer
 4 years ago
Best ResponseYou've already chosen the best response.1\[(3+3j)^3*(1j)^4=27(1+j)^3*(1j)^4=27*((1+j)(1j))^3(1j)\]\[=27*(1j^2)^3(1j)=27*(1+1)^3(1j)=27*2^3(1j)\]\[=27*8(1j)=216(1j)\]

asnaseer
 4 years ago
Best ResponseYou've already chosen the best response.1ok, so now we are left with:\[((3+j3)^{3} *(1j)^{4})/(1+j \sqrt{3})^{9}=\frac{216(1j)}{(1+j\sqrt{3})^9}\]

asnaseer
 4 years ago
Best ResponseYou've already chosen the best response.1next lets try and simplify the denominator...

asnaseer
 4 years ago
Best ResponseYou've already chosen the best response.1\[(1+j\sqrt{3})^2=1+j2\sqrt{3}+3j^2=1+j2\sqrt{3}3=2+j2\sqrt{3}=2(1j\sqrt{3})\]therefore:\[(1+j\sqrt{3})^3=(1+j\sqrt{3})^2(1+j\sqrt{3})=2(1j\sqrt{3})(1+j\sqrt{3})\]\[=2(1j^23)=2(1+3)=2*4=8\]

asnaseer
 4 years ago
Best ResponseYou've already chosen the best response.1therefore:\[(1+j\sqrt{3})^9=((1+j\sqrt{3})^3)^3=(8)^3=512\]

asnaseer
 4 years ago
Best ResponseYou've already chosen the best response.1can you complete the rest now?

EarthCitizen
 4 years ago
Best ResponseYou've already chosen the best response.0uhmm.., cpould you plz divide the top by the bottom for a final solution ?

asnaseer
 4 years ago
Best ResponseYou've already chosen the best response.1\[((3+j3)^{3} *(1j)^{4})/(1+j \sqrt{3})^{9}=\frac{216(1j)}{(1+j\sqrt{3})^9}=\frac{216(1j)}{512}=\frac{27(1j)}{64}\]

EarthCitizen
 4 years ago
Best ResponseYou've already chosen the best response.0kwl, your awesome!

asnaseer
 4 years ago
Best ResponseYou've already chosen the best response.1yw  the main point however is that you understand the process  I hope you did :)

asnaseer
 4 years ago
Best ResponseYou've already chosen the best response.1the "key" I guess is spotting how to get the equation into a form like: \((a+jb)(ajb)\) so that it can be simplified to \((a+jb)(ajb)=a^2+b^2\)

asnaseer
 4 years ago
Best ResponseYou've already chosen the best response.1that will come with practice.

EarthCitizen
 4 years ago
Best ResponseYou've already chosen the best response.0yh, hang on. so you used difference of two squares only ?

asnaseer
 4 years ago
Best ResponseYou've already chosen the best response.1\[a^2b^2=(a+b)(ab)\]\[a^2+b^2=(a+jb)(ajb)\]

EarthCitizen
 4 years ago
Best ResponseYou've already chosen the best response.0yh, so why didn't you use de Moivre's theorem ?

asnaseer
 4 years ago
Best ResponseYou've already chosen the best response.1you could, but then you would need to find the correct angle for:\[e^{j\theta}=\cos(\theta)+j\sin(\theta)\]and I /feel/ the approach I used instead is often much simpler.

EarthCitizen
 4 years ago
Best ResponseYou've already chosen the best response.0yh. a lot simpler coz it started getting cloudy , thanx tho

asnaseer
 4 years ago
Best ResponseYou've already chosen the best response.1np  you are more than welcome. I'm glad I was able to help.
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.