Complex numbers question (1 Viewer)

[TuQuoi]

Calculate the first six terms of z_n=(z_n-1)³-0.4i for
a. z₁=0.8
b. z₁=1

I'm not sure what the 'trick' to this question is. It's simple enough to find z₂, but I'm confused how to continue after that without having to manually calculate each term.

 

[KingOfActing]

I don't think there's any trick, it just looks like a silly calculation question made to make you hate the maths syllabus

 

[Paradoxica]

I don't think there's any trick, it just looks like a silly calculation question made to make you hate the maths syllabus

Not sure why they would put in something akin to the mandelbrot set but meh.

 

[dan964]

Calculate the first six terms of z_n=(z_n-1)³-0.4i for
a. z₁=0.8
b. z₁=1

I'm not sure what the 'trick' to this question is. It's simple enough to find z₂, but I'm confused how to continue after that without having to manually calculate each term.

write z_n in the form x+iy and likewise for z_(n-1) and equate real and imaginary parts.

 

[eyeseeyou]

How do you do this:

Solve the equation, express in the solutions a+bi where a and b are real for
z^2+(2+i)z+(2-2i)=0

(1+costheta+isintheta)/(1-costheta-isintheta)

 

[Paradoxica]

write z_n in the form x+iy and likewise for z_(n-1) and equate real and imaginary parts.

It's going to be fairly tedious, there's no way around it.

 

[InteGrand]

How do you do this:

Solve the equation, express in the solutions a+bi where a and b are real for
z^2+(2+i)z+(2-2i)=0

(1+costheta+isintheta)/(1-costheta-isintheta)

Just generally we can always solve these quadratic equations using the quadratic formula (you would generally need to compute the square root of a complex number to use this). Sometimes there are faster ways though.

For the last one, have you tried multiplying top and bottom by the conjugate of the denominator and using trig. identities?

 

[Paradoxica]

How do you do this:

Solve the equation, express in the solutions a+bi where a and b are real for
z^2+(2+i)z+(2-2i)=0

(1+costheta+isintheta)/(1-costheta-isintheta)

Observe that (-2-2i) + (i) = -2-i = -b

Further observe that (-2-2i)(i) = 2-2i

The sum and product of roots are known, so by inspection, the roots are z = -2(1+i), z = i

 

[Paradoxica]

Just generally we can always solve these quadratic equations using the quadratic formula (you would generally need to compute the square root of a complex number to use this). Sometimes there are faster ways though.

E.g. My way.

 

[eyeseeyou]

Just generally we can always solve these quadratic equations using the quadratic formula (you would generally need to compute the square root of a complex number to use this). Sometimes there are faster ways though.

For the last one, have you tried multiplying top and bottom by the conjugate of the denominator and using trig. identities?

I wasn't sure of the second one (of how to multiply the conjugate)

Do I multiply it by 1+costheta+isintheta or what?

 

[Paradoxica]

How do you do this:

Solve the equation, express in the solutions a+bi where a and b are real for
z^2+(2+i)z+(2-2i)=0

(1+costheta+isintheta)/(1-costheta-isintheta)

Observe that the expression is re-writable as (1+z)/(1-z) where z=cosθ+isinθ

Divide both parts by √z

Then you have (√z¯ +√z)/(√z¯ - √z) = Re(√z)/(-iIm(√z)) = itan(θ/2)

 

[eyeseeyou]

Thank you Paradoxica and Integrand