P6 Complex numbersWatch
Points P and Q represents the complex numbers w and z respectively in the Argand diagram. If w=[(1+z)/(Z+i)] and w=u +iv, express u and v in terms of x and y. Prove that when P describes a portion of the imaginary axis between the points representing i and -i, Q describes the whole of the positive imaginary axis.
Let z* denote the conjugate of z
w = (1+z)/(z+i) = (1+z)(z*-i)/(1+zz*)
= (x+y+x^2+y^2)/(1+x^2+y^2) + i(-1-x-y)/(1+x^2+y^2)
take real and imaginary parts to get u and v
We are then told that x = 0 and -1<y<1
But I am afraid that the question as written is wrong.
For example, if x = 0 and y = 1/2 we have the point
which is definitely not on the imaginary axis.
In fact the point w will map out a half-line of y=x-1
u + iv = [1+i(x+iy)]/[(x+iy)+i]
= [(1-y+ix)(x-i(y+1))]/[x²+(y+1)²] --- (Multiplying nom and denom by the conjugate of z.)
= [(1-y)x - i(1+y)(1-y) + ix² + x(y+1)]/[x²+(y+1)²]
Equate real and imaginary parts:
u = [x(1-y) + x(1+y)]/[x²+(y+1)²] = 2x/[x²+(y+1)²]
v = [-(1+y)(1-y) + x²]/[x²+(y+1)²] = (x² + y² - 1)/[x²+(y+1)²]
We know that u=0 and -1<v<1. So:
u = 2x/[x²+(y+1)²] = 0
v = (y² + x² - 1)/[x²+(y+1)²] = (y²-1)/(y+1)² = [(y-1)(y+1)]/(y+1)² = (y-1)/(y+1)
v < 1
(y-1)/(y+1) < 1
y-1 < y+1
-1 < 1, which is obvious. (From this we can conclude that y+1>0, since the < sign wasn't reversed when I multiplied both sides and the inequality still held.)
v > -1
(y-1)/(y+1) > -1
y-1 > -y-1 --- (Since y+1>0, i.e. the > doesn't become <.)
2y > 0
y > 0
Therefore, we have:
x = Re(z) = 0
y = Im(z) > 0
So Q describes the positive imaginary axis, as required.