((3+j3)^3*(1-j)^4)/((1+j(3)^1/2)^9)

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((3+j3)^3*(1-j)^4)/((1+j(3)^1/2)^9)

Mathematics
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\[((3+j3)^{3} *(1-j)^{4})/(1+j \sqrt{3})^{9}\]
yo
Just digesting that equation :)

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Other answers:

ok - it would be best to try and break it down a bit to simplify it first...
\[(3+3j)^3=(3(1+j))^3=3^3(1+j)^3=27(1+j)^3\]
next simplification...
\[(3+3j)^3*(1-j)^4=27(1+j)^3*(1-j)^4=27*((1+j)(1-j))^3(1-j)\]\[=27*(1-j^2)^3(1-j)=27*(1+1)^3(1-j)=27*2^3(1-j)\]\[=27*8(1-j)=216(1-j)\]
make sense so far?
yep
ok, so now we are left with:\[((3+j3)^{3} *(1-j)^{4})/(1+j \sqrt{3})^{9}=\frac{216(1-j)}{(1+j\sqrt{3})^9}\]
next lets try and simplify the denominator...
alryt
\[(1+j\sqrt{3})^2=1+j2\sqrt{3}+3j^2=1+j2\sqrt{3}-3=-2+j2\sqrt{3}=-2(1-j\sqrt{3})\]therefore:\[(1+j\sqrt{3})^3=(1+j\sqrt{3})^2(1+j\sqrt{3})=-2(1-j\sqrt{3})(1+j\sqrt{3})\]\[=-2(1-j^23)=-2(1+3)=-2*4=-8\]
therefore:\[(1+j\sqrt{3})^9=((1+j\sqrt{3})^3)^3=(-8)^3=-512\]
can you complete the rest now?
uhmm.., cpould you plz divide the top by the bottom for a final solution ?
\[((3+j3)^{3} *(1-j)^{4})/(1+j \sqrt{3})^{9}=\frac{216(1-j)}{(1+j\sqrt{3})^9}=\frac{216(1-j)}{-512}=-\frac{27(1-j)}{64}\]
kwl, your awesome!
yw - the main point however is that you understand the process - I hope you did :)
the "key" I guess is spotting how to get the equation into a form like: \((a+jb)(a-jb)\) so that it can be simplified to \((a+jb)(a-jb)=a^2+b^2\)
that will come with practice.
yh, hang on. so you used difference of two squares only ?
sum of two squares
\[a^2-b^2=(a+b)(a-b)\]\[a^2+b^2=(a+jb)(a-jb)\]
yh, so why didn't you use de Moivre's theorem ?
you 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.
yh. a lot simpler coz it started getting cloudy , thanx tho
np - you are more than welcome. I'm glad I was able to help.

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