Various qs on complex numbers (1 Viewer)

[Estel]

Firstly, a non-complex one (I've got a mindblank...)
how do you simplify cos(2(tan^-1(1/2)) without resort to calculator...

z = kcisx
prove that arg(z+k) =x/2

Find the locus of Re(z(z_bar+2)) = 3
[I got (-3,0) and (1,0)... bah]

|z-i| = Im(z)... describe the locus.
I got a parabola... Please confirm/find it to be wrong

Find (-1+rt3i)(1+i)
Hence, or otherwise find cos(11pi/12)... hence?!? How would you use the above?

Thanks...

 

[Mill]

(i)

Draw up a triangle. Let theta = invtan(1/2). I call theta @ from now on.

Then cos2@ = 2cos^2 @ - 1

Answer: 3/5

(ii)

Working on it. BBL.

(iii)

Let z = x+iy.

Answer: Circle. Centre (-1,0) Radius 2.

(iv)

Let z = x + iy.

Answer: Circle. Centre (0, 3/2) Radius root(5)/2.

(v)

Realise it by multiplying through by 1+i/1+i.

Answer: [ root(3) - 1 + i { root(3) + 1 } ] / 2

Without doing the actual calculations, I'd hazard the guess that cos(11pi/12) = Re { Cis 11pi/12 } = [ root(3) - 1 ] / 2.

 

[Constip8edSkunk]

1st q, let u = arctan 1/2, so u have cos2u , which is cos^2u - sin^2u.... but tanu = 1/2, so sinu = 1/sqrt5 and cosu = 2/sqrt5.... sub it in to get the answer,

2nd q, think of it geometrically, u are adding 2 complex vectors w/ lengths of k, along with the resulting z+k vector and the origin u get a rhombus, the diagonal(which is the resulting vector) of a rhombus bisects the angles of a rhombus(in this case x)

3rd q Re(z(conj(z)+2)) = 3 => Re(|z|^2 +2z) = 3 => 2Re(z) +|z|^2 = 3.... which would translate to x^2 + 2x + y^2 = 3

4th q |z-i| = Im(z) translates to x^2 + (y-1)^2 = y^2 , or x^2 + -2y +1 =0

5th q, the 1st complex no. is 2cis(2pi/3), the 2nd complex no. is sqrt2cis(pi/4).... so by multiplying them together u get sqrt8cis(11pi/12).... or sqrt8 ( cos(11pi/12) + isin(11pi/12)), so just equate real and imaginary parts to get desired answer


edit: corrected q4

 

[Mill]

Yeah I'm wrong on (v). Thought it was division.

 

[Estel]

Thankyou both very much!

Looks easy when you have the answer there =/

 

[nit]

For ii), you can also sub z=kcisx into LHS of what uve gotta prove...then it's just use of cos(2theta) formulae

 

[Ogden_Nash]

4th q |z-i| = Im(z) translates to x^2 + (y-1)^2 = y , or x^2 +y^2 -3y +1 =0

Hmm, isn't |z - i| = [x² + (y-1)²]^1/2 ?
Therefore x² + (y-1)² = y², giving a parabola with vertex (0,1/2) and directrix x = 0.

I may be wrong, can anyone confirm this at all? Thanks.

 

[VieTz_88]

hey isnt (iv) a parabola not a circle?

 

[mervvyn]

It looks like a parabola just from inspection: distance from a point (focus) is equal to distance to a line... i think.
I forget, is Im(z) y or iy?

 

[Ogden_Nash]

Im(z) = y.
Yeah, after looking at the question for a while, it's definitely a parabola.

 

[Mill]

Yeah, you guys are right. We missed the square root.

 

[Constip8edSkunk]

sorry, corrected