Trigonometry

[HCN86]

The number of cos60° + isin60° with respect to (-1/2)-( (?3)/2 )i

A. A square root
B. A cube root
C. A fourth root
D. A fifth root
E. None

Do I test out cos60° + isin60° or (-1/2)-( (?3)/2 )i ?

Sorry, I am not keen on math terminology "with respect to"

 

[tkhunny]

Start with deciding the "cis()" notation for the other complex number.

-1/2 and -sqrt(3)/2 Isn't that cis(240º)? Now what?

 

[HCN86]

I understand all of that but do I test cos 60 to see which n root would give me cos 270

or do I test cos 270 to see with n root give me cos 60.

 

[soroban]

Hello, HCN86!

$\displaystyle \text{The number }\cos60^o + i\sin60^o\text{ with respect to }\,\text{-}\tfrac{1}{2}-\tfrac{\sqrt{3}}{2}i \;\;\; {\bf[1]}$

. . $\displaystyle \text{(A) a square root} \qquad \text{(B) a cube root} \qquad \text{(C) a fourth root} \qquad \text{(D) a fifth root} \qquad \text{(E) None}$

$\displaystyle \text{We have: }\;z \;=\;\cos60^o + i\sin60^o \;=\;\cos\tfrac{\pi}{3} + i\sin\tfrac{\pi}{3}$


$\displaystyle \text{If }z\text{ is a square root of }{\bf[1]}\text{, then }z^2\text{ must equal }{\bf[1]}.$

. . \(\displaystyle \text{DeMoivre's Theorem: }\;z^2 \;=\;\cos2(\tfrac{\pi}{3}) + i\sin2(\tfrac{\pi}{3}) \;=\;\cos\tfrac{2\pi}{3} + i\sin\tfrac{2\pi}{3} \;=\; \text{-}\tfrac{1}{2} + i\tfrac{\sqrt{3}}{2} \;\hdots\;\text{ no}\)


$\displaystyle \text{If }z\text{ is a cube root of }{\bf[1]}\text{, then }z^3\text{ must equal }{\bf[1]}.$

. . \(\displaystyle \text{DeMoivre's Theorem: }\;z^3 \:=\:\cos3(\tfrac{\pi}{3}) + i\sin3(\tfrac{\pi}{3}) \;=\;\cos\pi + i\sin\pi \;=\;\text{-}1 \;\hdots\;\text{no}\)


$\displaystyle \text{If }z\text{ is a fourth root of }{\bf[1]}\text{, then }z^4\text{ must equal }{\bf[1]}.$

. . \(\displaystyle \text{DeMivre's theorem: }\;z^4 \;=\;\cos4(\tfrac{\pi}{3}) + i\sin4(\tfrac{\pi}{3}) \;=\;\cos\tfrac{4\pi}{3} + i\sin\tfrac{4\pi}{3} \;=\; \text{-}\tfrac{1}{2} - i\tfrac{\sqrt{3}}{2}\;\hdots\;\boxed{\text{Yes!}}\)


$\displaystyle \text{If }z\text{ is a fifth root of }{\bf[1]}\text{, then }z^5\text{ must equal }{\bf[1].}$

. . \(\displaystyle \text{DeMoivre's Theorem: }\;z^5 \;=\;\cos5(\tfrac{\pi}{3}) + i\sin5(\tfrac{\pi}{3}) \;=\;\cos\tfrac{5\pi}{3} + i\sin\tfrac{5\pi}{3} \;=\;\tfrac{1}{2} - \tfrac{\sqrt{3}}{2}\;\hdots\;\text{no}\)


$\displaystyle \text{Therefore: \;(C) }z\text{ is a fourth root.}$

 

[tkhunny]

I understand all of that

That was all you needed to solve the problem, excepting a little more effort.

Do it both ways and figure it out, rather than making soroban do it so you'll never have to make the effort.