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)
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