Roots of Complex Numbers (1 Viewer)

[kev-kun]

Hi all, bit confused on how to answer because my teacher rushes a LOT! Any help would be great!

<img src="http://latex.codecogs.com/gif.latex?$1)\&space;Find\&space;the\&space;fourth\&space;root\&space;of\&space;1&plus;i\sqrt3&space;\\$" title="$1)\ Find\ the\ fourth\ root\ of\ 1+i \sqrt3 \\$" />

<img src="http://latex.codecogs.com/png.latex?\\2)\(a)If\&space;w\&space;is\&space;a\&space;complex\&space;root\&space;of\&space;1.\&space;Find\&space;w\&space;in&space;\&space;polar&space;\&space;form&space;\&space;and\&space;show\&space;that\\&space;\indent(i)\&space;i&plus;w&plus;w^2=0\\&space;\indent(ii)\&space;(w^2)^2=0&space;\\&space;(b)Using\&space;these\&space;properties\&space;of&space;\&space;w,\&space;show\&space;that\\&space;\indent(i)(1-w)(1-w^2)(1-w^4)(1-w^5)=9&space;\\&space;\indent&space;(ii)\&space;(1&plus;2w&plus;3w^2)(1&plus;2w^2&plus;3w)=3&space;\\&space;\indent&space;\(iii)\&space;(1&plus;w)(1&plus;2w)(1&plus;3w)(1&plus;5w)=21" title="\\2)\(a)If\ w\ is\ a\ complex\ root\ of\ 1.\ Find\ w\ in \ polar \ form \ and\ show\ that\\ \indent(i)\ i+w+w^2=0\\ \indent(ii)\ (w^2)^2=0 \\ (b)Using\ these\ properties\ of \ w,\ show\ that\\ \indent(i)(1-w)(1-w^2)(1-w^4)(1-w^5)=9 \\ \indent (ii)\ (1+2w+3w^2)(1+2w^2+3w)=3 \\ \indent \(iii)\ (1+w)(1+2w)(1+3w)(1+5w)=21" />

 

[bottleofyarn]

For the first part, it looks like you use de Moivre's theorem after converting to polar form. Polar form then is the form.

I believe the second part means that w is a root of unity ie but it's rather confusing. Are you sure you've got it down right, because if then which contradicts with how w is a complex root of 1 (in part 2aii). If you get a question you don't quite get, it usually helps to work backwards.

 

[kev-kun]

Ohh okay thanks for the help! And for the second part I meant to say (w^2)^2=w. My mistake...

 

[bottleofyarn]

No problem! Ah, that makes sense then, which means w^3 = 1 ie w is the cube root of unity.

 

[hit patel]

yes thats what I was wondering but yep now that makes a lot more sense.