answer this please
this
yes that
\[-4 -4\sqrt{3}i \]
did you try it on the calculator?
-4-6.9 i
thats as simplified as it gets :)
express it in trigonometric form :((
youve gotta answer somethin first lol @ilikephysics2
x+iy = r cos theta + i r.sin theta.
i can't relate with you because we haven't taken that during our discussion :((
so u have x and y r= \(\sqrt{x^2+y^2}\) \(\huge\theta=tan^{-1}\frac{y}{x}\)
how is that?
8cis240degrees
do u know how to find magnitude of complex numbers ?
no:((
can you explain it ? :)
magnitude of x+iy is \(\sqrt{x^2+y^2}\) and magnitude of r cos theta + i r.sin theta is r.
i found the value of r which is 8 and what will i do next??
theta is 60 , what to do next?
u write it in the form of r (cos theta + i sin theta).
how will i arrive in the answer 8cis240degrees??
u have the answer ?
yes.. but i don't know how will i arrive in that answer..
8cis240degrees is certainly not the answer......
what actual answer u have ?
that is what the higher year gave to me, but he didn't show me how to solve it and that's why im asking this.. the possible answers are: a. 8cis120 b. 8cis240 c. 8cis150 d. 8cis300
so the question was to find REAL PART of that ??
and thhat is COS not CIS
the question is this: express -4-4(square root of)3 i in trigonometric form.. i guess the letter "i" means it is imaginary and that is why it is written CIS?? but i just guess that..
u got 8, right?...now u calculated theta as tan inverse of root 3, right ? u got 60. but, tan inverse of root 3 is also 240 so 8cos240
may be its the short of 8(cos 240 + i sin 240)
i don't know :)) but thank you for the answer :))
remember those conversion formulas.........
yes i will :))
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