if arg [(z-2)/(z-6i)]=π /2, show that x^2+y^2-2x-6y=0 2. z=2-√3i, find the real value of q such as arg(z^2+qz)=-π /3?

1.
Since arg = π/2 ,
( z - 2 ) / ( z - 6i ) = 0 + ki , for some real number k
z - 2 = zki + 6k
z ( 1 - ki ) = 6k + 2

z
= ( 6k + 2 ) / ( 1 - ki )
= ( 6k + 2 )( 1 + ki ) / [ ( 1 - ki )( 1 + ki ) ]
= [ ( 6k + 2 ) + k( 6k + 2 )i ] / ( 1 + k^2 )

So,
x = ( 6k + 2 ) / ( 1 + k^2 )
y = k( 6k + 2 ) / ( 1 + k^2 ) = kx


x^2 + y^2 - 2x - 6y
= x^2 + k^2*x^2 - 2x - 6kx
= x [ ( 1 + k^2 )x - ( 2 + 6k ) ]
= x [ ( 6k + 2 ) - ( 2 + 6k )]
= 0

Q.E.D.

2.
z^2 + qz
= 4 + 3i^2 - 4√3i + q( 2 - √3i )
= 1 - 4√3i + 2q - q√3i
= ( 1 + 2q ) - √3( 4 + q )i

tan( - π / 3 ) = y / x
- √3 = -√3( 4 + q ) / ( 1 + 2q )
4 + q = 1 + 2q
q = 3 ... Ans

arg[(z - 2)/(z - 6i)] = π/2..........z = x + iy


(z - 2)/(z - 6i) = ik

(z - 2) = ik(z - 6i)

z - 2 = ikz + 6k

z(1 - ik) = 6k + 2

z = (6k + 2)/(1 - ik)

　= (6k + 2)(1 + ik)/(1 + k²)

　= [(6k + 2) + i(6k² + 2k)]/(1 + k²)


x = (6k + 2)/(1 + k²) , y = (6k² + 2k)/(1 + k²)

x - 1 = [(6k + 2) - (1 + k²)]/(1 + k²)

　　 = (6k + 2 - 1 - k²)/(1 + k²)

　　 = (1 + 6k - k²)/(1 + k²)

y - 3 = (6k² + 2k)/(1 + k²) - 3

　　　= (6k² + 2k - 3 - 3k²)/(1 + k²)

　　　= (3k² + 2k - 3)/(1 + k²)　
　

　x² + y² - 2x - 6y

= (x - 1)² + (y - 3)² - 10

= (1 + 6k - k²)²/(1 + k²)² + (3k² + 2k - 3)²/(1 + k²)² - 10

= (1 + 36k² + k⁴ + 12k - 12k³ - 2k²　+　9k⁴ + 4k² + 9 + 12k³ - 12k - 18k²)/(1 + k²)² - 10

= (10 + 20k² + 10k⁴)/(1 + 2k² + k⁴) - 10 = 0



z = (2 - i√3) ..............z² = (1 - i⋅4√3)

arg(z² + qz) = - π/3

z² + qz = k(1 - i√3)

(1 - i⋅4√3) + q(2 - i√3) = k(1 - i√3)

1 - i⋅4√3 + 2q - iq√3 = k - ik√3

(1 + 2q - k) + i⋅√3( - 4 - q + k) = 0

　1 + 2q - k = 0
- 4 - q + k = 0

- 3 + q = 0................q = 3

a) Let z=x+yi, then [(z-2)/(z-6i)]={(x^2+y^2-2x-6y)+[xy+(x-2)(y-6)]i}/[x^2+(y-6)^2].
argument=pi/2=>
(x^2+y^2-2x-6y)/[x^2+(y-6)^2]=cos(pi/2)=>
x^2+y^2-2x-6y=0

(b) z=2-sqrt(3)i=>
z^2+qz=(7+2q)-(4+q)sqrt(3)i
argument=(-pi/3)=>
(7+2q)/sqrt{(7+2q)^2+[(4+q)^2]/3}=
cos(-pi/3)=1/2=>
35q^2+244q+425=0=>
q= -3.4 or q= -3.571429 (rejected)

Check: [7+2(-3.4)]/sqrt{[7+2(-3.4)]^2+
[(4+(-3.4)]^2/3}=0.5=cos(-pi/3)

1) Geometrically z moves in a a way that ZP(2 , 0) and ZQ(0, 6) are always at right ∠s, i.e., on a circle centre mid-point of PQ - the point (1, 3) - and radius 0.5√(2² + 6²) = √10. Its eqn is (x - 1)² + (y - 3)² = 10 or x² + y² - 2x - 6y = 0.

2) Lopez's solution for q is spot on - I can't better it.