^{3}

Solution:

(1 – i)

^{3}= (1- i)(1- i)(1- i) = -2 - 2i

Then arg(z) = arctan (y/x) = arctan (-2/-2) = arctan (1) = p/ 4

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

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

Recall:

**=**

*r exp(i q )*

*cos(q ) + isin(q )***p/ 4) = Arg (z)**

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

Arg(z1) + Arg(z2)

Show it doesn’t hold if :

z1 = (-1 + iÖ3) and z2 = -(Ö3) + i

arg(z1) = Arg(z1) = arctan (Ö3/ -1) = arctan (- Ö3)= (p/ 3)

arg (z2) = Arg (z2) = arctan (1/ -(Ö3) ) = arctan (- Ö3/ 3)

Doing the working:

Arg(z1) + Arg(z2) = arctan (Ö3/ -1) + arctan (1/ -(Ö3) ) = (p/ 3 + (-p/ 6)) = p/ 6

But: Arg(z1) Arg(z) = arctan [(Ö3/ -1) (1/ -(Ö3) ]= arctan (1 ) = p/ 4

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

Solution:

Arg(z) = arg(z) = arctan (-1/ Ö3) = arctan (-Ö3/3) = -p/ 6

which satisfies the requirement for the principal value such that:

-p < arg z < p, (and r = 2)

Then: ln (z) = ln(r) + i(q) = ln(2) + i(- p/6)

But ln 2 = 0.693, so:

ln(z) = ( 0.693) + (-i) 3.14/6 = 0.693 - 0.523i

5. Find Arg(z) for:

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

Then: Arg (z) = arg (z) = 7p/ 3

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

Note: z = exp (2 + ip)

Then Arg(z) = arctan (y/x) = arctan (p/ 2) = arctan (1.57) = 57.3 deg or 1 rad (radian)

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