Question 16 Complex Number on Trial Exam (2 Viewers)

[bobbybrownsenior]

Can anyone check my working out and help me for this "complex" question hahaa sorry.
(a) Find all roots of z^4-1 = 0. Draw their positions on an Argand diagram. (easy)
(b) (hard) Find the solutions of the equation (z+1)^4 = 4(z-1)^4

Ok I started off doing it like this (correct me if I'm wrong):
(z+1)^4 = 4(z-1)^4
(z+1 / z-1) ^4 = 4

Let z+1/z-1 = w
w^4 = 4

Do I solve this like equation w^4 = 4 and find all of its roots in a+ib form and equation it with (z+1/z-1) ...? I'm a little confused at this bit, can someone explain? Thank you

 

[Trebla]

Can anyone check my working out and help me for this "complex" question hahaa sorry.
(a) Find all roots of z^4-1 = 0. Draw their positions on an Argand diagram. (easy)
(b) (hard) Find the solutions of the equation (z+1)^4 = 4(z-1)^4

Ok I started off doing it like this (correct me if I'm wrong):
(z+1)^4 = 4(z-1)^4
(z+1 / z-1) ^4 = 4

Let z+1/z-1 = w
w^4 = 4

Do I solve this like equation w^4 = 4 and find all of its roots in a+ib form and equation it with (z+1/z-1) ...? I'm a little confused at this bit, can someone explain? Thank you

Express (z+1)/(z-1) in mod-arg form by applying some half angle formulas which remove the 1's. It then becomes similar to the standard roots of unity problem but the modulus and arguments are slightly different.

 

[Drsoccerball]

Express (z+1)/(z-1) in mod-arg form by applying some half angle formulas which remove the 1's. It then becomes similar to the standard roots of unity problem but the modulus and arguments are slightly different.

Yep The final expression is

 

[Drsoccerball]

Yep The final expression is

apparently not ahhaha

 

[braintic]

I don't see any need to use trig.

Just write (z+1)²=2(z-1)² and (z+1)²=-2(z-1)².

Then rearrange and solve each equation using the quadratic formula.


What is more interesting is that the answers all lie on a circle centred at (3,0) with radius 2√2.
I'm wondering if anyone can show why that must be the case without actually solving. (I can't)

 

[wiliaamnguuyen]

I gott z= +-1 +-i as an answer. 4 answers total for z, is that correct?

 

[bobbybrownsenior]

Yes, that's correct! I'm going to try Trebla's method, that certainly looks very interesting way to solve it

 

[braintic]

I gott z= +-1 +-i as an answer. 4 answers total for z, is that correct?

They are NOT the answers.
Two answers are purely real. And all answers include a √2.
The answers above don't work on substitution.

 

[wiliaamnguuyen]

Oh oops, how do you do it then since they aren't the answers?? Curious

 

[braintic]

Oh oops, how do you do it then since they aren't the answers?? Curious

See my post 5 before this one.

 

[bobbybrownsenior]

I still don't understand this :'(, Braintic, do I just expand and rearrange to solve like normal quadratics??

 

[braintic]

I still don't understand this :'(, Braintic, do I just expand and rearrange to solve like normal quadratics??

A quadratic is a quadratic. It doesn't matter that is has resulted from the solution of a quartic - it is still solved the same way.
You haven't had practice at solving quadratics which give complex solutions?

 

[glittergal96]

I don't see any need to use trig.

Just write (z+1)²=2(z-1)² and (z+1)²=-2(z-1)².

Then rearrange and solve each equation using the quadratic formula.


What is more interesting is that the answers all lie on a circle centred at (3,0) with radius 2√2.
I'm wondering if anyone can show why that must be the case without actually solving. (I can't)

More generally, suppose you are solving for w^n=1 (1), where w=(az+b)/(cz+d)=T(z) for some complex a,b,c,d with ad-bc not equal to zero (*).

Such maps T are called "fractional linear transformations" on the complex plane, and have lots of nice properties. One of these properties that is easily provable with MX2 knowledge is that T maps generalised circles to generalised circles. (A generalised circle is a term that encompasses both circles and lines, as the latter behaves in many ways like a circle of infinite radius.)

We know that all solutions w to (1) lie on the unit circle. So all solutions z must be such that T(z) lies on the unit circle.

Given condition (*) though, fractional linear transformations are invertible, and their inverse is also a fractional linear transformation.

So all solutions z lie on the generalised circle that is the image of the unit circle under the fractional linear transformaton T^(-1).

 

[glittergal96]

More generally, suppose you are solving for w^n=1 (1), where w=(az+b)/(cz+d)=T(z) for some complex a,b,c,d with ad-bc not equal to zero (*).

Such maps T are called "fractional linear transformations" on the complex plane, and have lots of nice properties. One of these properties that is easily provable with MX2 knowledge is that T maps generalised circles to generalised circles. (A generalised circle is a term that encompasses both circles and lines, as the latter behaves in many ways like a circle of infinite radius.)

We know that all solutions w to (1) lie on the unit circle. So all solutions z must be such that T(z) lies on the unit circle.

Given condition (*) though, fractional linear transformations are invertible, and their inverse is also a fractional linear transformation.

So all solutions z lie on the generalised circle that is the image of the unit circle under the fractional linear transformaton T^(-1).

A fun MX2 exercise could be to calculate the radius and centre of the generalised circle that the unit circle gets mapped to by T in terms of the coefficients of T.