Need help with question! (1 Viewer)

[alcronin]

I'm working through the Cambridge 4 Unit maths book and it doesn't have worked solutions one question is:

Z = 1 + i sqrt3. Find the smallest positive integer 'n' for which z^n is real and evaluate z^n for this value of n. Show that there is no integral value of n for which z^n is imaginary.

Any help would be appreciated or a link to worked solutions! Thanks

 

[bobmcbob365]

If a complex number is real, then the argument of the complex number will be equal to 0, pi, 2pi, ...
Use De Moivre's Theorem as well.

 

[HeroicPandas]

If z^n is REAL, then IM(z)=0, ie. when you put z into mod-arg form, you say since z^n is real therefore its imaginary parts are zero, then solve for n

 

[HeroicPandas]

I give up on part 2 xD

 

[alcronin]

Thanks for the help guys I got all the way to z^n = 2^n (cos n pi/3) then I have trouble going further?

 

[bobmcbob365]

 

[HeroicPandas]

Thanks for the help guys I got all the way to z^n = 2^n (cos n pi/3) then I have trouble going further?

Wheres cis lol

 

[alcronin]

Well I thought the imaginary part had to equal 0 so sin pi/3 would have to equal 0? So it would just drop out of the equation??

 

[HeroicPandas]

Well I thought the imaginary part had to equal 0 so sin pi/3 would have to equal 0? So it would just drop out of the equation??

Don't worry about my post, look at bob's professional answers- he used a better method

 

[alcronin]

Oh that makes sense! Thanks bob, I didn't pick up that z^n = k pi and thats what stumped me. Thanks you both!

 

[alcronin]

By the way, does anyone have the worked solutions for Cambridge 4 unit?

 

[bobmcbob365]

Lol don't worry. Your method is equally as correct. As long as you get the answer it's all good. =p

 

[bobmcbob365]

no problem.

hold on, I think there's a link somewhere on these forums.

 

[alcronin]

Oh so I could have continued with z^n = 2^n (cos n pi/3)?? and thanks

 

[bobmcbob365]

http://community.boredofstudies.org/showthread.php?t=226576

 

[HeroicPandas]

Lol don't worry. Your method is equally as correct. As long as you get the answer it's all good. =p

ok, i like ur method a lot! using arguments, my method i gotta bring out general solutions for sine or the way i call it "Fundamental Formula"

 

[HeroicPandas]

Oh so I could have continued with z^n = 2^n (cos n pi/3)?? and thanks

No, when u put in mod-arg, u get z^n = 2^n[ cos(npi/2) +isin(npi/2)]

 

[HeroicPandas]

No, when u put in mod-arg, u get z^n = 2^n[ cos(npi/2) +isin(npi/2)]

If z^n = real, then Im(z)=0 (expand in the 2^n first)

THAT IS: 2^n [sin(npi/2)] =0

get me?

EDIT: ALL imaginary PARTS are equal to zero, so we ignore REAL parts(no "i")

Imaginary parts: Stuff infront of "i" xD

 

[alcronin]

Oh ok so then we get 2^n (sin n pi/3) = 0 right? then solve that somehow?

 

[GoldyOrNugget]

Oh you guys are gonna love me.

Cambridge fully worked solutions: https://dl.dropbox.com/u/1103246/4U answers.zip