Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2011 / Paper 1 / 17 / a

The point has positive real part and negative imaginary part. Its complex modulus is

1 + 3 = 2

, and its principal complex argument is

- π /3

. Hence

∣ z ∣ = 2, ar g z = - \frac{π}{3} + 2 πk, k \in Z .

ii

e^{iπ /2} = i

and

2 e^{iπ /4} = 1 + i

. Thus

z = 1 + 2 i

lies in the first quadrant, giving

∣ z ∣ = 5, ar g z = arctan (2) + 2 πk, k \in Z .

Its principal argument is approximately

1.10715

Write

1 + i = 2 e^{iπ /4}

and multiply the phases:

z = 2 e^{i (π /4 + π /6)} = 2 e^{5 πi /12} .

Therefore

∣ z ∣ = 2, ar g z = \frac{5 π}{12} + 2 πk, k \in Z .

The displayed representative already lies in the principal argument interval.

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