Question

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

Answer 1

\[((3+j3)^{3} *(1-j)^{4})/(1+j \sqrt{3})^{9}\]

Answer 2

yo

Answer 3

Just digesting that equation :)

Answer 4

ok - it would be best to try and break it down a bit to simplify it first...

Answer 5

\[(3+3j)^3=(3(1+j))^3=3^3(1+j)^3=27(1+j)^3\]

Answer 6

next simplification...

Answer 7

\[(3+3j)^3*(1-j)^4=27(1+j)^3*(1-j)^4=27*((1+j)(1-j))^3(1-j)\]\[=27*(1-j^2)^3(1-j)=27*(1+1)^3(1-j)=27*2^3(1-j)\]\[=27*8(1-j)=216(1-j)\]

Answer 8

make sense so far?

Answer 9

yep

Answer 10

ok, so now we are left with:\[((3+j3)^{3} *(1-j)^{4})/(1+j \sqrt{3})^{9}=\frac{216(1-j)}{(1+j\sqrt{3})^9}\]

Answer 11

next lets try and simplify the denominator...

Answer 12

alryt

Answer 13

\[(1+j\sqrt{3})^2=1+j2\sqrt{3}+3j^2=1+j2\sqrt{3}-3=-2+j2\sqrt{3}=-2(1-j\sqrt{3})\]therefore:\[(1+j\sqrt{3})^3=(1+j\sqrt{3})^2(1+j\sqrt{3})=-2(1-j\sqrt{3})(1+j\sqrt{3})\]\[=-2(1-j^23)=-2(1+3)=-2*4=-8\]

Answer 14

therefore:\[(1+j\sqrt{3})^9=((1+j\sqrt{3})^3)^3=(-8)^3=-512\]

Answer 15

can you complete the rest now?

Answer 16

uhmm.., cpould you plz divide the top by the bottom for a final solution ?

Answer 17

\[((3+j3)^{3} *(1-j)^{4})/(1+j \sqrt{3})^{9}=\frac{216(1-j)}{(1+j\sqrt{3})^9}=\frac{216(1-j)}{-512}=-\frac{27(1-j)}{64}\]

Answer 18

kwl, your awesome!

Answer 19

yw - the main point however is that you understand the process - I hope you did :)

Answer 20

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 \((a+jb)(a-jb)=a^2+b^2\)

Answer 21

that will come with practice.

Answer 22

yh, hang on. so you used difference of two squares only ?

Answer 23

sum of two squares

Answer 24

\[a^2-b^2=(a+b)(a-b)\]\[a^2+b^2=(a+jb)(a-jb)\]

Answer 25

yh, so why didn't you use de Moivre's theorem ?

Answer 26

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.

Answer 27

yh. a lot simpler coz it started getting cloudy , thanx tho

Answer 28

np - you are more than welcome. I'm glad I was able to help.