Now, Noting That Χ = Θ + Π/2 Or Χ Π/2 = Θ, We Have by Substitution & Noting Ei(Π/2) = I

In problem 9.7, I will draw a picture of 2 cases & derive two very similar results. First, the picture to work with will be considered case 1 i

real

-i χ

-il θ

2π ●z = a – ib

The center as given in the book is –il/2π and z will be a point on the ever growing outer circle, say a – ib. Chi χ will remain as the arc length on the unit circle ≅ to the same full angle that leads to z so that χ = θ + π/2; this following the description of χ on pg 167. real of z = a = l/2πcos(θ) & the imaginary of z = b = l/2πsin(θ) + l/2π; so that

z = l/2πcos(θ) – i(l/2πsin(θ) + l/2π)

= l/2π[(cos(θ) – isin(θ)] – il/2π

= l/2π(e–iθ – i)

Now, noting that χ = θ + π/2 or χ – π/2 = θ, we have by substitution & noting ei(π/2) = i

z = l/2π(e–i(χ –π/2) – i)

= l/2π(e–i(χ)·ei(π/2) – i)

= l/2π(e–i(χ)·i– i)

= i l/2π(e–i(χ) – 1) which is the desired derivation.

Case 2: (Only the description of χ i

is changed)

real

-i

-il θ

2π ●z = a – ib

z will still be as described above i.e. z = a – ib. However, Chi χ will be the arc length on the same growing circle as z is on; so that χ = l/2π(θ + π/2). Hence, the same description for z:

z = l/2πcos(θ) – i(l/2πsin(θ) + l/2π)

= l/2π[(cos(θ) – isin(θ)] – il/2π

= l/2π(e–iθ – i)

But, now noting that χ = l/2π(θ + π/2) or 2π/l(χ) – π/2 = θ, we have by similar substitution & algebraic manipulation:

z = l/2π(e–i(2π/l(χ) –π/2) – i)

= l/2π(e–i2π/l(χ)·e–i(π/2) – i)

= l/2π(e–i2π/l(χ)·i– i)

= –i l/2π(e–i2π/l(χ) – 1)

= –iω(e–iωχ – 1) (where ω = 2π/l)

which is a similar result just that χ is the arc length on the circle whose circumference l.