numerator
a(a+bi-i) - 1(a+bi-i) -bi(a+bi-i)
or
a^2+abi-ai-a-bi+i-abi+b^2-b
or
a^2 -a -ai + b^2 -b -bi + i
if that is imaginary then the real part of the numerator
a^2 - a + b^2 - b = 0 which is your step 3
let z= a+bi be a complex number.
it is given that the quotient (z-i)/(z-1) is purely imaginary.
show that z lies on a circle and determine the centre and radius of this circle.
OK, here's my working out...( i am using "X" as my times symbol)
step1:
(a+bi-i)/(a+bi-1)X[(a-1)-bi]/[(a-1)-bi]
step 2:
=[aa-a-abi+(b-1)(a-1)i+b.b-b]/[(a-1)(a-1)+bb]
and i got stuck here...
i have checked the answer and the next step is: step 3:aa+bb-a-b=0
and the circle centres at (0.5,0.5)
radius is 1/root 2.
but i don't really get it. how to you get to step 3 from step 2 and how do you get the radius and the centre point of the circle from step 3??
thanks!
P.S:i know this is looking pretty messy...sorry...
4 answers
if a circle then of form
(a-h)^2 + (b-k)^2 = R^2 center at (h,k)
here
(a-.5)^2 + (b-.5)^2 = ????
a^2 -a +.25 +b^2 -b +.25 = ???
BUT a^2-a +b^2-b = 0 step 3
so
.25 + .25 = .5
.5 = R^2 = 1/2
so R = 1/sqrt 2
(a-h)^2 + (b-k)^2 = R^2 center at (h,k)
here
(a-.5)^2 + (b-.5)^2 = ????
a^2 -a +.25 +b^2 -b +.25 = ???
BUT a^2-a +b^2-b = 0 step 3
so
.25 + .25 = .5
.5 = R^2 = 1/2
so R = 1/sqrt 2
thank you so much!!
I don't quite get it. How can we substitute the values of h and k as .5 since we have to prove it first?
1 answer