The Student Room Group
Uni students - Complete our survey >>
Uni students - Complete our survey >>
Uni students - Complete our survey >>

Complex numbers question

(i) Find the roots of the equation z^2 + (2 sqrt 3)z +4 = 0, giving your answers in the form x + iy, where x and y are real.
(ii) State the modulus and argument of each root.
(iii) Showing all your working, verify that each root also satisfies the equation
z^6 = −64.

How do I do the third question without expanding the roots of z to the power 6?
Original post by bmqib
(i) Find the roots of the equation z^2 + (2 sqrt 3)z +4 = 0, giving your answers in the form x + iy, where x and y are real.
(ii) State the modulus and argument of each root.
(iii) Showing all your working, verify that each root also satisfies the equation
z^6 = −64.

How do I do the third question without expanding the roots of z to the power 6?


Do you know that Mod(z^6)=(modz)^6 and that arg(z^6) = 6x(arg z)?
(or put another way, De Moivre's theorem)
z6+64=(z2+4)(z4−4z2+16)z^6+64=(z^2+4)(z^4-4z^2+16)

Assuming OP doesn't know De Moivre. Split the quartic into the product of two quadratics - you know what one of them is going to be.
(edited 12 years ago)
Reply 3
Avatar for bmqib
bmqib
OP
13
Original post by tiny hobbit
Do you know that Mod(z^6)=(modz)^6 and that arg(z^6) = 6x(arg z)?
(or put another way, De Moivre's theorem)


Yeah but that's not in the curriculum so there could be a different way to do it... do I have to find z in the r(sin theta + i cos theta) form then?
Reply 4
Avatar for bmqib
bmqib
OP
13
Original post by Mr M
z6+64=(z2+4)(z4+4z2+16)z^6+64=(z^2+4)(z^4+4z^2+16)

Assuming OP doesn't know De Moivre. Solve the disguised quadratic.


But how do I factorise it that quickly? and the right hand side is not equal to lhs?
(edited 12 years ago)
Original post by bmqib
But how do I factorise it that quickly? and the right hand side is not equal to lhs?


It is now I corrected the wrong sign!

http://www.purplemath.com/modules/specfact2.htm
Reply 6
Avatar for Azland
Azland
8
Original post by bmqib
(i) Find the roots of the equation z^2 + (2 sqrt 3)z +4 = 0, giving your answers in the form x + iy, where x and y are real.
(ii) State the modulus and argument of each root.
(iii) Showing all your working, verify that each root also satisfies the equation
z^6 = −64.

How do I do the third question without expanding the roots of z to the power 6?

Lol, I just did the same paper myself. Wanted to post the same question. CIE sucks.

Quick Reply

Related discussions

Latest

Trending

Trending

Articles for you