The Mobius transformation T(z)=(az+b)/(z-1) maps unit circle onto the line Re(z)=3 in such a way that the interior of the circle is mapped to the left of the line and that the origin gets mapped to 2+i. Find a.
To find the value of 'a' in the Mobius transformation T(z) = (az + b) / (z - 1), we can use the known properties of Mobius transformations.
We are given that the Mobius transformation maps the unit circle onto the line Re(z) = 3. This means that any point on the unit circle will be transformed to a point on the line Re(z) = 3.
First, let's consider a point on the unit circle, z = e^(iθ), where θ represents the angle. We can express this point as z = cos(θ) + isin(θ).
Now, let's apply the Mobius transformation T(z) to this point:
T(z) = (az + b) / (z - 1)
Substituting z = e^(iθ), we have:
T(e^(iθ)) = (ae^(iθ) + b) / (e^(iθ) - 1)
We know that this point on the unit circle gets mapped to a point on the line Re(z) = 3.
Since the line Re(z) = 3, this means that the real part of T(z) should be 3.
Therefore, Re(T(e^(iθ))) = 3.
Let's find Re(T(e^(iθ))):
Re(T(e^(iθ))) = Re((ae^(iθ) + b) / (e^(iθ) - 1))
To simplify this expression, multiply the numerator and the denominator by the complex conjugate of e^(iθ) - 1:
Re(T(e^(iθ))) = Re(((ae^(iθ) + b) / (e^(iθ) - 1)) * ((e^(-iθ) - 1) / (e^(-iθ) - 1)))
Simplifying further, we have:
Re(T(e^(iθ))) = Re((a(e^(iθ)e^(-iθ)) + b(e^(-iθ) - 1)) / (e^(iθ)e^(-iθ) - 1))
Since e^(-iθ)e^(iθ) = 1, we can simplify the expression to:
Re(T(e^(iθ))) = Re((a + b(e^(-iθ) - 1)) / (1 - 1))
This becomes:
Re(T(e^(iθ))) = Re(a + b(e^(-iθ) - 1))
Finally, since we want Re(T(e^(iθ))) to be 3 for any point on the unit circle, we can set Re(T(e^(iθ))) = 3 and solve for 'a'.
3 = Re(a + b(e^(-iθ) - 1))
Since Re(a + b(e^(-iθ) - 1)) = a, we have:
3 = a
Therefore, 'a' is equal to 3.
So, the value of 'a' in the Mobius transformation T(z) = (az + b) / (z - 1) is 3.