Problem: Find all the roots of: (-2 + 2i)^1/3

First we find the amplitude so that:

r = [8^1/2]^1/3 = [8]^1/6 = (2)^1/2

The argument (theta) is: theta =arctan (y/x) = arctan(1) = pi/4

(Note here that one must take care to specify the choice of arctan(y/x) so the point corresponding to r and (theta) lies in the appropriate quadrant. The reason for this is that tan(theta) has period pi while sin(theta) and cos(theta) has period 2pi. Thus, we specific arctan(y/x) in quadrant 1)

So, the first root is:

w0= (2)^1/2[cos (pi/4) + isin(pi/ 4)] = (2)^1/2[(2)^1/2/2 + i(2)^1/2/2]

The second root (k=1) is:

w1 = (2)^1/2[cos(pi/ 4 + 2 pi /3) + isin(pi/ 4 + 2 pi /3)]

= (2)^1/2 [cos(11 pi/12)) + isin(11 pi/12))]

The third root is:

w2 = (2)^1/2[cos(pi/ 4 + 4 pi /3) + isin(pi/ 4 + 4 pi/3)]

=(2)^1/2[cos(19pi/12) + isin(19 pi/12)]

And, of course, the specific root numbers can be worked out as shown in the last instalment though this isn’t really needed once the angles are given.

Now, to the Principal Value:

For a given complex number z not equal 0, the value of arg(z) that lies in the range:

-pi < theta < pi is called the “principal value of arg(z)” and is denoted Arg(z). Thus:

Arg(z) = theta where -pi < theta < pi. The relation between arg(z) and Arg(z) can be set out:

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

EXAMPLES:

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

We can use basic algebra of complex numbers to obtain:

z = 2/ (i-1) = -1 –i

then arg(z) = arctan(1) = pi/ 4

since we can let k=0 then arg(z) = Arg(z) = pi/4

2- Find Arg(z) for z = 1 –i

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

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

3- Find Arg(z) for (-1-i(3)^1/2)^2

z = (1-i(3)^1/2)^2 = -2 + 3.464i

so arg(z)= arctan(-3.464/2) = arctan (-(3)^1/2)) = (2pi /3) = Arg(z)

4- Calculate the principal value of ln(z) when z = 1 +i

Arg(z) = arg(z) = pi/ 4 which satisfies the requirement for the principal value such that: -pi < arg z < pi, and r = (2)^1/2

Then: ln (z) = ln(r) + i(theta) = ln(r) + i(pi/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 –pi located on real axis) so the principal value of arg(z) is:

Arg(z) = pi and ln(z) = ln (abs(z)) + pi i = ln(abs(5)) + pi i

Note:

*abs*denotes "absolute value of".

## No comments:

Post a Comment