MYTUTOR SUBJECT ANSWERS

1081 views

Factorise z^3+1 into a linear and quadratic factor. Let y=(1+i√3)/2. Show that y is a cube root of -1. Show that y^2=y-1. Find the value of (1-y)^6.

To factorise z3+1 into a linear and quadratic factor, it is best to try the factor theorem first. We set the equation to zero: z3+1=0.

We need to find a value of z such that the above is true. By inspection we can see that if z=-1 then the above is true because (-1)3+1=0 since (-1)3=-1.

Therefore, (z+1) is a factor. The other factor is quadratic. We know, by inspection, that the quadratic factor will look like: (z2+kx+1) where k is a constant. This is because z3+1 can be written as a product of its factors: z3+1=(z+1)(z2+kx+1). The coefficient of the z3 term and constant term are 1 so we know the coefficient of the z2 and constant term in the quadratic. By inspection we can also see that k=-1.

The second part of the questions asks us to show that y3=-1. There are several approaches for this. You can square y to obtain y2 then multiply it by y again to obtain y3 and show that it is equal to -1. You can also write y in polar form, then apply De Moivre’s theorem to obtain y3 then convert back into Cartesian form. The final method is as follows:

We want to show that y3=-1. Earlier we wrote: z3=-1 which leads to z3+1=(z+1)(z2-z+1)=0. We know that if y is a cubic root of -1 then it cannot be equal to -1, so for the above to be true then (z2-z+1) must be equal to 0. We simply solve the quadratic (z2-z+1)=0 using the quadratic formula. The formula gives us that z=(1+i√3)/2 after taking the positive result. We note that this is equal to y. Since y=x then y is a cube root of -1.

The third part of the question asks us to show that y2=y-1. Using our working from earlier, we know that y is a root of z2-z+1=0 so y2-y+1=0. Rearranging this we have y2=y-1 as required. Alternatively you can find y2  then show that it is equal to y-1.

The final part of the question is to find the value of (1-y)6. Instinctively most people would use the binomial expansion but there is a far more elegant solution. From the previous part we have y2=y-1. If we multiply both sides by -1 then:

(-y2 )=1-y

Raising both sides to the power 6 as the question required and using power laws:

(-y2)6=(1-y)6

Why would we do this though? Maybe you can see why; there is a piece of information from earlier that we can use, if not, keep following and you will see that this method is very clever.

(-y)6*2=(y)12=(1-y)6

The negative sign disappears as squaring a number makes it positive.

We can manipulate this further:

(y)12=(y)3*4=(y3)4

Earlier we had (y3)=-1 and so ultimately the value of (1-y)6=(-1)4=1.

Henry L. IB Maths tutor, A Level Maths tutor, 13 plus  Maths tutor, 1...

11 months ago

Answered by Henry, an IB Maths tutor with MyTutor


Still stuck? Get one-to-one help from a personally interviewed subject specialist

29 SUBJECT SPECIALISTS

£26 /hr

Tadas T.

Degree: MMathPhil Mathematics and Philosophy (Bachelors) - Oxford, St Anne's College University

Subjects offered:Maths, Further Mathematics + 3 more

Maths
Further Mathematics
.MAT.
-Personal Statements-
-Oxbridge Preparation-

“University of Oxford Maths and Philosophy student happy to help students learn and stay motivated!”

£30 /hr

Szymon K.

Degree: Mechanical Engineering (Masters) - Manchester University

Subjects offered:Maths, Physics+ 1 more

Maths
Physics
Further Mathematics

“About me: I am a student pursuing my degree in Mechanical Engineering at the University of Manchester. Ever since I can remember I have always been passionate about passing my knowledge to others. As a tutor I taught people of differen...”

£30 /hr

Ravi S.

Degree: Mechanical Engineering (Masters) - Bristol University

Subjects offered:Maths, Physics

Maths
Physics

“Masters Mech eng. student, University of Bristol. Did Physics, Maths + Further Maths. Not free till 19th may due to exams”

About the author

Henry L.

Currently unavailable: for regular students

Degree: Chemical Engineering (Masters) - Bath University

Subjects offered:Maths, -Personal Statements-

Maths
-Personal Statements-

“About me- I am a chemical engineering student at the University of Bath. Mathematics and the sciences are subjects I have always enjoyed and I especially love applying the knowledge I have learnt in all sorts of areas, such as cooking ...”

You may also like...

Posts by Henry

Factorise z^3+1 into a linear and quadratic factor. Let y=(1+i√3)/2. Show that y is a cube root of -1. Show that y^2=y-1. Find the value of (1-y)^6.

Make y the subject of (y/x)+(2y/(x+4))=3

Other IB Maths questions

differentiate: y^2 + 3xy + x + y = 8

When integrating by parts, how do I decide which part of the integrand is u or f(x) and which dv or g'(x)?

How do you calculate the probability P(X < x) for a normally distributed random variable X?

Why is (-1)*(-1)=1?

View IB Maths tutors

We use cookies to improve your site experience. By continuing to use this website, we'll assume that you're OK with this. Dismiss

mtw:mercury1:status:ok