Given that arg((z-1)/(z-2i)) = pi/4 describe the locus of P and sketch this on an argand diagram. Show that the point (1,3) lies on this locus.
I know it is an arc subtended from the points (0,2) and (1,0) but I do not know how to find the centre of the circle which I am guessing is how I would go on to draw the diagram accurately and answer the second part about the point (1,3)
To find the locus of points P on the complex plane with an argument of pi/4 for the expression (z-1)/(z-2i), we can start by expressing z in terms of a complex variable. Let's call this complex variable a + bi, where a and b are real numbers.
Now, let's substitute z = a + bi into the expression (z-1)/(z-2i):
(z-1)/(z-2i) = (a + bi - 1)/(a + bi - 2i)
Next, we need to rewrite this expression in a + bi form by multiplying the numerator and denominator by the conjugate of the denominator:
= [(a + bi - 1)(a - bi + 2i)]/[(a - 2i)(a + bi - 2i)]
Expanding and simplifying the numerator:
= (a^2 - 2ai + 2ai - b^2i^2 - a + abi + 2bi - 2ibi + 2i^2)/[(a - 2i)(a + bi - 2i)]
Since i^2 = -1:
= (a^2 + 2abi - b^2 - a + abi + 2bi + 2)/(a^2 + bi^2 - 2ai - 2bi - 2a + 4i^2)
= (a^2 + b^2 + 1 + (3a + ab + 2b)i)/(a^2 + b^2 - 2a + 2b - 4)
We know that the argument of this expression is pi/4. So, we can equate the imaginary part to the real part multiplied by tan(pi/4) to get:
3a + ab + 2b = (a^2 + b^2 + 1)tan(pi/4)
Since tan(pi/4) = 1, the equation becomes:
3a + ab + 2b = a^2 + b^2 + 1
Rewriting this equation in the quadratic form:
a^2 - (3 - b)a + (b^2 - 2b + 1) = 0
Now, we have the equation of a curve on the complex plane that represents the locus of points P with the argument of pi/4 for the expression (z-1)/(z-2i). It is a curve given by the quadratic equation above.
To determine whether the point (1,3) lies on this locus, we can substitute a = 1 and b = 3 into the equation:
1^2 - (3 - 3)1 + (3^2 - 2(3) + 1) = 0
1 - 0 + (9 - 6 + 1) = 0
1 + 4 = 0
5 ≠ 0
Since the equation does not hold true, the point (1,3) does not lie on the locus.
To accurately sketch the locus on an Argand diagram, you can solve the quadratic equation graphically or numerically to find the roots and plot them as points on the complex plane.