Question 1207779
<pre>
Here's another way to do the (a) part:

Since  |z| = |w| = 1

This problem is about points on the unit circle, where the horizontal axis
is the real axis and the vertical axis is the imaginary axis.

{{{drawing(200,200,-1.2,1.2,-1.2,1.2,graph(200,200,-1.2,1.2,-1.2,1.2),circle(0,0,1), line(0,0,.6427876097,.7660444431),line(0,0,.6427876097,-.7660444431),
red(arc(0,0,.5,-.5,0,50)), green(arc(0,0,.5,-.5,310,3600))


)}}}  


All complex numbers on the unit circle have modulus 1., the radius of the
unit circle.

The conjugate of any complex number on the unit circle is obviously its
reflection in the horizontal (real) axis.  It is also obvious that the 
argument (angle) of the conjugate is the negative of the argument (angle).

The reciprocal of any complex number is that complex number when multiplied by
the complex number, gives the result 1, which is the complex number 1+0i,
thought of as the point (1,0).

DeMoivre's theorem tell us that to multiply two complex numbers, we
multiply their moduli and add their arguments.  

So when we multiply the arguments of any two complex numbers on the unit circle,
we'll get 1.
 
It's also obvious to see that when you add the arguments, the
red and green arcs above, you will get 0 for the argument.   

So it's obvious that the reciprocal and conjugate of any complex number of the
unit circle are the same. 

This is what the (a) part asks you to prove:   

We can write z and w in trigonometric form:

{{{1(cos(theta)+i*sin(theta))}}}

{{{cos(theta)+i*sin(theta)}}}

We will show that the reciprocal equals the conjugate:

{{{1/(cos(theta)+i*sin(theta))}}}

{{{1/(cos(theta)+i*sin(theta))}}}{{{""*""}}}{{{(cos(theta)-i*sin(theta))/(cos(theta)-i*sin(theta))}}}

{{{(cos(theta)-i*sin(theta))/(cos^2(theta)+sin^2(theta))}}}

Since the denominator equals 1, 

{{{1/(cos(theta)-i*sin(theta))}}}{{{""=""}}}{{{cos(theta)+i*sin(theta)}}}

So the reciprocal of a complex number on the unit circle is also its
conjugate.

Edwin</pre>