Imaginay Nos (3 Viewers)

[Lukybear]

Question on Complex Nos.



Show that

 

[Lukybear]

Also

Find x in the domain 0<x<pi/2

 

[ninetypercent]

cube both sides

(x+iy) = X^3 + 3X^2(iY) + 3X(iy)^2 + (iy)^3 = X^3 + 3X^2iY - 3XY^2 -iY^3

equating real and imaginary parts

 

[Cazic]

Also

Find x in the domain 0<x>

I can't make sense of the question, but if it's just asking you solve for x then we have:

 

[Lukybear]

What about

Divide
x^3-2-2i by x+1-i

 

[gurmies]

Sorry, that was meant to be:

 

[Lukybear]

Thxs Gurmies, much appreciated.

Ive got more tho.

 

[untouchablecuz]

cube both sides

(x+iy) = X^3 + 3X^2(iY) + 3X(iy)^2 + (iy)^3 = X^3 + 3X^2iY - 3XY^2 -iY^3

equating real and imaginary parts

(this method is abit quicker, good work either way 90%)

 

[Lukybear]

I dont get this question at all.

Prove that if the ratio is purely imaginary, the point z lies on the circle whose centre is at the point 1/2(1+i) and whose radius is 1/sqrt 2

Shouldnt it it mean the ratio = Z (upper case) (since z-1/z-i is a complex no.), and that the point Z lies on the circle...

Because if it is the lower case z, then rearranging the ratio would be necessary, unless im wrong?

 

[Lukybear]

Also this question has me stunned.

Two points, P, Q represent the complex no.s z, 2z+3+i respectively, where P moves on the circle |z|=k how does Q move?

 

[khorne]

Not 100% this is right, but:

let p be a point on the circle |z|=k.

Mod. p = k
Arg p = @

2z + 3+i = 2z + (3+i)

so in a sense you are multiplying z by 2 and adding 3+i

Since the argument of 2z and z are the same, |2z| = 2k

I.e 2z would just move as a circle with twice the radius.

3+i however, moves the circle down ,i and to the left, 3, so the origin is at (-3,-i) and its radius is 2k.

 

[untouchablecuz]

I dont get this question at all.

Prove that if the ratio is purely imaginary, the point z lies on the circle whose centre is at the point 1/2(1+i) and whose radius is 1/sqrt 2

Shouldnt it it mean the ratio = Z (upper case) (since z-1/z-i is a complex no.), and that the point Z lies on the circle...

Because if it is the lower case z, then rearranging the ratio would be necessary, unless im wrong?

im sure you can find the equation of the locus of now (which is indeed a circle)

 

[cyl123]

Close but not quite.... cos adding 3+i onto the vector 2z should move it upwards by i and right by 3, so logically the centre of the circle should shift by the same amount

Or convert into parametrics question:
Let P=z=x+iy so x^2+y^2=k^2 (x,y,X,Y are real)

Let Q=X+iY such that Q=2z+3+i
So equating gives X=2x+3, Y=2y+1
So 2x=X-3 , 2y=Y-1
So since 4x^2+4y^2=4k^2
then (X-3)^2+(Y-1)^2=4k^2
where coordinates of Q is (X,Y) and Q moves in a circle with centre (3,1) and radius 2k

Not 100% this is right, but:

let p be a point on the circle |z|=k.

Mod. p = k
Arg p = @

2z + 3+i = 2z + (3+i)

so in a sense you are multiplying z by 2 and adding 3+i

Since the argument of 2z and z are the same, |2z| = 2k

I.e 2z would just move as a circle with twice the radius.

3+i however, moves the circle down ,i and to the left, 3, so the origin is at (-3,-i) and its radius is 2k.

 

[khorne]

my bad, I forgot I was working with vectors still

 

[Lukybear]

More to go:

, show that when z describes the circle |z| = 1, completely in one direction, then w describes the circle |w|=1 competely in the other direction.

 

[Lukybear]

Just on this one, when using parametrics, as so stated, did anyone get two varaible for one value? If then, how do you solve, since i havent done parametrics yet.

 

[Trebla]

More to go:

, show that when z describes the circle |z| = 1, completely in one direction, then w describes the circle |w|=1 competely in the other direction.

Not sure about the "other direction" part:

 

[Trebla]

 

[Lukybear]

I do observe my mistake Trebla. I do very indeed thank you, for being such a good person, helping me on a monday morning.

 

[cutemouse]

Sorry, that was meant to be:

Or you could use long division...