How do I solve arg(z−iz+i)=π4?

Question

sirpsta3u · Accepted Answer

z−iz+i=x+iy−ix+iy+i=x2+(y−1)2∠tan−1y−1xx2+(y+1)2∠tan−1y+1x  So arg (z−iz+i)=tan−1y−1x−tan−1y+1x=π4  taking tangents of both sides  y−1x−y−1x1+y−1xy−1x=1  This simplifies to x2+2x+y2=1 or (x+1)2+y2=2

Stella Calderon · Accepted Answer

First multiply top and bottom by the complex conjugate of the denominator

\frac{z + i}{z - i} = \frac{(z + i) (\overset{―}{z} + i)}{(z - i) (\overset{―}{z} + i)}

= \frac{z \overset{―}{z} + i (z + \overset{―}{z}) - 1}{z \overset{―}{z} + i (z - \overset{―}{z}) + 1}

If

z = x + i y then \overset{―}{z} = x - i y, z + \overset{―}{z} = 2 x, z - \overset{―}{z} = 2 i y, z \overset{―}{z} = x^{2} + y^{2}

and the fraction simplifies to

\frac{z + i}{z - i} = \frac{x^{2} + y^{2} - 1 + i 2 x}{x^{2} + y^{2} + 1 - 2 y}

As the denominator is positive real this will not change the argument of the fraction. Now if

a r g (w) = \frac{π}{4} then w = r (1 + i)

for some

r > 0

. So the real and imaginary parts of the numerator must be equal, and both positive, that is

x^{2} + y^{2} - 1 = 2 x

. Rearrange

x^{2} - 2 x + 1 + y^{2} = 2 and {(x - 1)}^{2} + y^{2} = 2

a circle center (1,0) with radius

\sqrt{2}

.
There might be a neater way of showing this, the formula is related to Möbius transformation. These transformations have the general form

\frac{a z + b}{c z + d}

and preserve angles, map every straight line to a line or circle, and map every circle to a line or circle. The inverse transformation will also map straight lines, like the line

a r g (w) = \frac{π}{4}

, to circles.

How do I solve arg(\frac{z-i}{z+i})=\frac{\pi}{4}?

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