Need Some Trigonometry Help
I'm on the de Moivre's theorm and polar coordinates part. I already know the answers, I just don't know how to get them
1.) Expressing your answer in rcis(x) form, where x is between zero and 360, solve x^(3)+3i=0
2.) If z=(-1/2)+(sqrt(3)/2)i, evaluate z^(0)+z^(1)+z^(2)+z^(3)+.......+z^(3002). Express your answer in a+bi form
December 8, 2012
6 Comments • Newest first
[quote=OriginalUser]Pre-Calculus has a chapter called Trigonometry.[/quote]
Sure, but complex roots != trig.
I came here ready with mah soh cah toa,
left with a toa cah soh HAA!
edit:
This is precalc, not trig? In any case, I've forgotten all of my precalc, but I believe you do something like, move the 3i to the right, and replace X with rcis(theta), where r is the right side squared, and theta starts at some initial value, increasing by 2pi/n for each other root.
[quote=charismatic]For (1) I don't see anything to solve. Also, I'm assuming you mean x^(3 + 3i), correct? Does that equal something, or are we just changing its form?[/quote]
It's suppose to equal 0, sorry. Also its just x cubed plus 3 i.
For (1), I believe you need to draw an Argand diagram (Where the imaginary axis is 'y' and the real axis is 'x') and you can practically translate it (though I can't be too sure - I've never been put in a position where 'x' is to the power of a number. It's usually in the form 'x + yi' where 'a' and 'b' are constants.)
I'm thinking you'd need to substitute 'z' into the equation (or possibly translate it into polar form from rectangular form.)