is called the “principal value of arg(z)” and is denoted Arg(z).

Thus: Arg(z) = q where -p < q < p .

The relation between arg(z) and Arg(z) can be set out:

arg(z) = Arg(z) + 2 p k (where k is an integer)

EXAMPLES:

1) Find the principal value for z = 2/ (i-1)

We can use the basic algebra of complex numbers to obtain: z = 2/ (i-1) = -1 –i then arg(z) = arctan(1) = p/ 4 since we can let k=0 then arg(z) = Arg(z) = p/4 .

2) Find Arg(z) for z = 1 –i

We have arg(z) = arctan(y/x) = arctan(-1) = -p /4

So: Arg(z) = arg(z) = -p /4

3) Find Arg(z) for (1 -iÖ3)

^{2}

z = (1 -iÖ3)

^{2}= -2 + 3.464i so arg(z)= arctan(-3.464/2)

= arctan (-Ö3) = (2p /3) = Arg(z)

4) Calculate the principal value of ln(z) when z = 1 +i and Arg(z) = arg(z) = p/ 4 which satisfies the requirement for the principal value such that:

-p < arg z < p, and r = Ö2

Then: ln (z) = ln(r) + i(q) = ln(r) + i(p /4)

But ln 2 = 0.693, so:

ln(z) = ½(0.693) + 3.14/4(i) = 0.347 + 0.785i

5) Find the principal value of (-5)

Here, z is a real negative (angle at –p located on the real axis) so the principal value of arg(z) is:

Arg(z) = p and ln(z) = ln êz + pi = ln(5) + pi

Problems for the Math Maven:

1. Find the principal value of (1 – i)

^{3}

2. Find Arg(z) if z = 2i exp[-i(3p/ 4)]

3. In general, the identity holds that: Arg(z1 z2) =

Arg(z1) + Arg(z2)

Show it doesn’t hold if :

z1 = (-1 + i Ö3) and z2 = -(Ö3) + i

4. Find the principal value for log(z) where z = Ö3 –i

5. Find Arg(z) for:

a) z = 8 [exp i(7p/ 3)]

b) z = exp(2) exp(i p)

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