Angle of a Complex Number

  • #1
Hi all.

Suppose I am looking for the following quantity: [itex]\sphericalangle[/itex] cn, where cn = [itex]\frac{sin(\frac{nπ}{2})}{nπ}[/itex]. cn is a complex number.

According to the book, "Signals and Systems" by Edward Kamen 2nd. Ed., [itex]\sphericalangle[/itex] cn = π for n = 3, 7, 11 ... , and cn = 0, for all other n.

The angle of a complex number is defined by arctan(b/a), where b is the imaginary component of the complex number, and a is the real component. In this case, there is no imaginary component (b).

Using the Euler Formula: e[itex]^{iθ}[/itex] = cos (θ) + jsin(θ), I can derive the relation of sin(θ) to an exponential, but I feel this is going backwards.

Any hints/insights is appreciated.

Answers and Replies

  • #2
I figured it out. This is a complex number with only a real part. e.g.:

cn = [itex]\frac{sin(\frac{nπ}{2})}{nπ}[/itex]

For even n, note that cn is equal to 0. The angle of a zero complex number is 0.

For n = 1, 5, 9, ..., c = a + bj, a is 1, b is 0. arctan(0/1) is 0 degrees.

For n = 3, 7, 11, ..., c = a + bj, a is -1, b is 0. 180 - arctan(0/1) is 180 degrees.
  • #3
Science Advisor
indeed, with complex numbers, the end-points of the unit semi-circle lie on the real axis. one feels intutively, that real numbers should correspond to "angle 0".

but if one regards a complex number in polar form, then the radius (norm, modulus) is positive, so to get the negative real numbers we need a non-zero angle (pi).

this peculiarity of the complex numbers, messes up things like the logarithm function and the square root function. for example, a square root should be a "half-angle formula".

but if our angle is near 2pi, our angle is also near 0, and we're going to have a "jump" (discontinuity) when taking square roots.

so complex numbers solve one problem (we can take more square roots), but give us another (we can't make square roots be near each other, for angles near each other on different sides of some "bad" angle).

if only the exponential wasn't periodic in i, right?