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martins
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Report Thread starter 16 years ago
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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.
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RichE
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(Original post by martins)
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*)

= (z*+zz*-iz-i)/(1+zz*)

= (x-iy+x^2+y^2-ix+y-i)/(1+x^2+y^2)

= (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

(1-2i)/3

which is definitely not on the imaginary axis.

In fact the point w will map out a half-line of y=x-1
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martins
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Report Thread starter 16 years ago
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Sorry guys, the transformation should have been w=[(1+zi)/(z+i)]
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dvs
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Report 16 years ago
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w = u + iv = (1+zi)/(z+i)

Let z=x+iy:
u + iv = [1+i(x+iy)]/[(x+iy)+i]
= (1-y+ix)/(x+i(y+1))
= [(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
=> x=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)

Case #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.)

Case #2:
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.
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