1A (Vectors and Matrices)

z = e^{i θ}

. Factoring out the half-angle gives

1 + z = e^{i θ /2} (e^{- i θ /2} + e^{i θ /2}) = 2 cos \frac{θ}{2} e^{i θ /2} .

Since

0 \leq θ < π

, the cosine is positive. Therefore

∣1 + z ∣ = 2 cos \frac{θ}{2}, ar g (1 + z) = \frac{θ}{2} .

On the Argand diagram, the points

0, 1, z, 1 + z

form a rhombus. Its diagonal from

0

1 + z

bisects the angle between the unit vectors

1

and

z

, so its direction is

θ /2

; resolving either side along the diagonal gives its length

2 cos (θ /2)

b

Similarly,

1 - z = e^{i θ /2} (e^{- i θ /2} - e^{i θ /2}) = - 2 i sin \frac{θ}{2} e^{i θ /2} = 2 sin \frac{θ}{2} e^{i (θ /2 - π /2)} .

Thus, for

0 < θ < π

∣1 - z ∣ = 2 sin \frac{θ}{2}, ar g (1 - z) = \frac{θ}{2} - \frac{π}{2} .

θ = 0

1 - z = 0

, whose complex argument is undefined. Geometrically,

∣1 - z ∣

is the chord joining the two unit-circle points

z

and

1

. The isosceles triangle with vertex angle

θ

has half-chord

sin (θ /2)

, and the chord direction is perpendicular to the radius through its midpoint, giving the stated argument.

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