Let C$ be the circle with center -3i and radius 2. The Mobius transformation R given by the equation R(z)=b/(z+d) maps C to the unit circle in such a way that the inside of C maps to the outside of the unit circle. Find |R(1-2i)|.
To find |R(1-2i)|, we need to apply the Mobius transformation R to the point 1-2i and find the modulus or absolute value of the resulting complex number.
First, let's substitute z = 1-2i into the equation R(z) = b/(z+d):
R(z) = b/(z+d)
R(1-2i) = b/((1-2i)+d)
Since we want R to map the circle C to the unit circle, we know that the point -3i on C must map to 1 on the unit circle. This gives us:
R(-3i) = b/((-3i)+d) = 1
Solving this equation for d, we get:
((-3i)+d) = b
d = b + 3i
Now, let's substitute z = 1-2i and d = b + 3i back into the equation R(z) = b/(z+d):
R(z) = b/(z+d)
R(1-2i) = b/((1-2i)+(b+3i))
Simplifying this expression, we get:
R(1-2i) = b/(1-2i+b+3i)
= b/(2+b+i)
So, |R(1-2i)| is the modulus or absolute value of b/(2+b+i).
Note that to find the modulus of a complex number, we can use the formula:
|a+bi| = √(a^2 + b^2)
In this case, our complex number is b/(2+b+i), so we need to find the real and imaginary parts of this complex number to calculate the modulus.
Let's multiply the numerator and denominator by the conjugate of the denominator to simplify the expression:
R(1-2i) = b/(2+b+i)
= b(2+b-i)/((2+b+i)(2+b-i))
= b(2+b-i)/(4+4b+b^2-i^2)
= b(2+b-i)/(4+4b+b^2+1)
= b(2+b-i)/(b^2 + 4b + 5)
The real part of b(2+b-i)/(b^2 + 4b + 5) is:
Re[R(1-2i)] = (2b + b^2)/(b^2 + 4b + 5)
The imaginary part of b(2+b-i)/(b^2 + 4b + 5) is:
Im[R(1-2i)] = -b/(b^2 + 4b + 5)
Therefore, the modulus of b/(2+b+i) is:
|R(1-2i)| = √[(Re[R(1-2i)])^2 + (Im[R(1-2i)])^2]
= √[((2b + b^2)/(b^2 + 4b + 5))^2 + (-b/(b^2 + 4b + 5))^2]
The final step is to substitute the value of b into the expression and evaluate it to get the answer.