complex... (1 Viewer)

[littleboy]

hey, can you guys help me out with these questions?

1. Determine the locus of the complex number z given arg (z-2) = arg ( z+2) + pi/4 . Sketch the locus on argand diagram

2. If q is real and z = (3+iq)/(3-iq) show that q varies the point in the complex plane which represents z lies on a circle. Find the centre and radius of this circle.

3. If x is real and (x+i)^4 is imaginary, find the possible values of x in surd form

 

[香港!]

hey, can you guys help me out with these questions?

1. Determine the locus of the complex number z given arg (z-2) = arg ( z+2) + pi/4 . Sketch the locus on argand diagram

2. If q is real and z = (3+iq)/(3-iq) show that q varies the point in the complex plane which represents z lies on a circle. Find the centre and radius of this circle.

3. If x is real and (x+i)^4 is imaginary, find the possible values of x in surd form

1.arg (z-2) = arg ( z+2) + pi/4
arg (z-2)-arg (z+2)=pi\4
arg [ (z-2)\(z+2) ] =pi\4
so locus would be half a circle drawn from 2 to -2 anti clockwise such that angle inscribed in it is pi\4

3. (x+i)^4=x^4+4x³i+6x²i²+4xi³+i^4
=x^4-6x²+1+i(....)
its imaginary so real parts=0
x^4-6x²+1=0
x²=[6+-sqrt(36-4)]\2
=[6+-4sqrt 2]\2
=3-2sqrt 2
.: x=+-sqrt (3-2sqrt 2)

haven't done dis stuff for long time... plz correct me if any is wrong

I don't get that q2 btw.. "q varies the point in the complex ...."

 

[Antwan23q]

very nice with question 3, but i dont get how the first one, i get the working, but dont u get an actualy locus equation? like x+y or something like that?

 

[香港!]

very nice with question 3, but i dont get how the first one, i get the working, but dont u get an actualy locus equation? like x+y or something like that?

dunno..
but in all the types of q's like this one that i've done...
the locus are just descriptions, i nvr saw an equation

 

[robbo_145]

Question 2

z = (3 + iq)²/(3-iq)(3+iq)
= (9 + 6qi - q²)/( 9 + q²)

z = (9-q²)/(9+q²) + 6qi/(9+q²)
let z = x + iy
x = (9-q²)/(9+q²)
y = 6qi/(9+q²)

the question says it should be a circle so square both sides

x² = (9-q²)²/(9+q²)²
x² = ((9+q²)² - 36q²)/(9+q²)²
x² = 1 - 36q²/(9+q²)²

y² = 36q²/(9+q²)²

∴ x² + y² = 1

 

[香港!]

for question 3 you have
=[6+-4sqrt 2]\2

and then the next line you have
=3-2sqrt 2

shouldn't it be "3+-2sqrt 2"

so the answer would be
x=+-sqrt (3+-2sqrt 2)

But if you use the "3+2sqrt 2", and put it in your calculator, it turns out to be not a solution. So I left it out...
hehe
But maybe im made a typing mistake.......

 

[sagarn88]

1.arg (z-2) = arg ( z+2) + pi/4
arg (z-2)-arg (z+2)=pi\4
arg [ (z-2)\(z+2) ] =pi\4
so locus would be half a circle drawn from 2 to -2 anti clockwise such that angle inscribed in it is pi\4

3. (x+i)^4=x^4+4x³i+6x²i²+4xi³+i^4
=x^4-6x²+1+i(....)
its imaginary so real parts=0
x^4-6x²+1=0
x²=[6+-sqrt(36-4)]\2
=[6+-4sqrt 2]\2
=3-2sqrt 2
.: x=+-sqrt (3-2sqrt 2)

haven't done dis stuff for long time... plz correct me if any is wrong

I don't get that q2 btw.. "q varies the point in the complex ...."

for question 3 you have
=[6+-4sqrt 2]\2

and then the next line you have
=3-2sqrt 2

shouldn't it be "3+-2sqrt 2"

so the answer would be
x=+-sqrt (3+-2sqrt 2)

 

[robbo_145]

using the circumference-centre angle property of a circle it can be shown the circle is
x² + (y-2)² = 8
(in the diagram W = π/4)

 

[Antwan23q]

Ive got some working out for question 3.

arg(z-2) - arg(z+2) =pi/4
arg [(z-2)/(z+2)] = pi/4

(z-2)/(z+2) = (x-2+iy)/(x+2+iy) then rationalising the denomiator

= (x-2+iy)(x+2-iy)/[(x+2)^2+y^2]

=[x^2-4+4iy]/[(x+2)^2+y^2]

now taking the seperating the real and imaginary

= (x^2+y^2-4)/[(x+2)^2+y^2] + 4yi/[(x+2)^2+y^2]

tan(pi/4) = {4y/[(x+2)^2+y^2]} / {(x^2+y^2-4)/[(x+2)^2+y^2]}

4y/(x^2+y^2-4) = 1

x^2+y^2-4=4y

x^2+(y-2)^2 = 8

 

[KFunk]

Also make sure you account for the fact that 0 < y &le; 2(1 + &radic;2)

 

[jake2.0]

using the circumference-centre angle property of a circle it can be shown the circle is
x² + (y-2)² = 8
(in the diagram W = &pi;/4)

would the circle be present under the x-axis?

 

[Antwan23q]

no, so it shouldnt have a line in that picture, and yes, kfunk right yet again