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...
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
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
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?
No problem! I'll guide you through the steps to better understand the solution.
Step 1: Simplifying the expression (z-i)/(z-1)
We want to simplify the expression (z-i)/(z-1), so let's start by applying the distributive property and simplifying the numerator and the denominator:
(a+bi - i) / (a+bi - 1)
= (a+bi - i) / (a - 1 + bi)
= [(a - 1) + (bi - i) ] / [(a - 1) + bi]
Step 2: Rationalizing the numerator and denominator
To rationalize the numerator, we multiply both the numerator and denominator by the conjugate of the denominator:
[(a - 1) + (bi - i) ] / [(a - 1) + bi] * [(a - 1) - bi] / [(a - 1) - bi]
Now, let's multiply out the numerator and denominator:
= [(a - 1)(a - 1) - (bi - i)(a - 1) + (bi - i)(a - 1) - (bi - i)(bi - i)] / [(a - 1)(a - 1) - (bi)(bi)]
= [(a - 1)(a - 1) - (bi - i)(a - 1) + (bi - i)(a - 1) - (bi - i)^2] / [(a - 1)^2 - (bi)^2]
= [a^2 - 2a + 1 - (a - 1)(bi - i) + (a - 1)(bi - i) - (bi)^2 +2bi - bi + i^2] / [(a - 1)^2 - b^2]
Next, simplify further:
= [a^2 - 2a + 1 - a(bi - i) + (bi - i) + a(bi - i) - b^2i^2 + 2bi - bi - 1] / [(a - 1)^2 - b^2]
= [a^2 - 2a + 1 - abi + ai + bi - i - 1 + b^2 + 2bi - bi] / [(a - 1)^2 - b^2]
Step 3: Simplifying the expression
Now, let's simplify the above expression further:
= [a^2 - 2a + 1 + (ai - i) + (bi + 1 - bi - 1) + b^2 + bi - bi] / [(a - 1)^2 - b^2]
= [a^2 - 2a + 1 + ai - i + b^2 + bi - bi] / [(a - 1)^2 - b^2]
= [a^2 - 2a + b^2 + 1 + ai - i] / [(a - 1)^2 - b^2]
= [a^2 - 2a + b^2 + 1 + (ai - i)] / [(a - 1)^2 - b^2]
Now, we see that the expression in the numerator is purely imaginary, which means the real part is zero. Therefore, we have:
a^2 - 2a + b^2 + 1 = 0
This is the equation of a circle!
Step 4: Center and radius of the circle
To find the center and radius of the circle, let's rewrite the equation in the form of (x - h)^2 + (y - k)^2 = r^2.
Rewrite it as:
(a - 1)^2 + b^2 = 1
Comparing this to the standard form, we can see that the center is (1, 0) and the radius is 1.
Therefore, the complex number z lies on a circle centered at (1, 0) with a radius of 1.
I hope this explanation clarifies the steps and helps you understand the solution better. Let me know if you have any further questions!