## Sunday, March 10, 2013

### Solutions: Principal Value Problems

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

arg (z) = (- 3p/ 4)  = Arg (z)

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)