SOLUTION: The complex numbers z and w satisfy |z| = |w| = 1 and zw is not equal to -1. (a) Prove that overline{z} = 1/z and overline{w} = 1/w. (b) Prove that (z + w)/(zw + 1) is a real

Algebra ->  Complex Numbers Imaginary Numbers Solvers and Lesson -> SOLUTION: The complex numbers z and w satisfy |z| = |w| = 1 and zw is not equal to -1. (a) Prove that overline{z} = 1/z and overline{w} = 1/w. (b) Prove that (z + w)/(zw + 1) is a real      Log On


   



Question 1189372: The complex numbers z and w satisfy |z| = |w| = 1 and zw is not equal to -1.
(a) Prove that overline{z} = 1/z and overline{w} = 1/w.
(b) Prove that (z + w)/(zw + 1) is a real number.

Answer by ikleyn(52784) About Me  (Show Source):
You can put this solution on YOUR website!
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The complex numbers z and w satisfy |z| = |w| = 1 and zw is not equal to -1.
(a) Prove that overline{z} = 1/z and overline{w} = 1/w.
(b) Prove that (z + w)/(zw + 1) is a real number.
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                    Part (a)


Part (a) is the widely known fact. Students learn it on the beginner steps of studying complex numbers.

The proof is very short and straightforward.


Notice, that in part (a), we should prove only first statement for z, since the statement for w is mathematically THE SAME.


    If z = a + bi, then overline(z) = a - bi.

    If |z| = 1, it means that sqrt%28a%5E2+%2B+b%5E2%29 = 1, or, which is the same, a^2 + b^2 = 1.


    Now,  1%2Fz = 1%2F%28a%2Bbi%29 = %281%2F%28a%2Bbi%29%29%2A%28%28a-bi%29%2F%28a-bi%29%29 = %28a-bi%29%2A%281%2F%28a%5E2%2Bb%5E2%29%29 = a-bi = overline(z), 

    and the statement is proved.


                    Part (b)


Part (b) is not widely known, which makes it interesting.


    Since |z| = 1 and |w| = 1, it means that z and w are the unit vectors of the length 1: their endpoints lie on the unit circle.

    To calculate (z+w), apply the parallelogram's rule.  Since the sides of the parallelogram on vectors z and w are equal,

    the parallelogram is a rhombus.  The sum (z+w) is the diagonal of the parallelogram, and since parallelogram is a rhombus,

    arg(z+w) is  EITHER %28arg%28z%29+%2B+arg%28w%29%29%2F2  OR  %28arg%28z%29+%2B+arg%28w%29%29%2F2+%2B+pi.   Here arg() means the argument of complex number.



    The first case   arg(z+w) = %28arg%28z%29+%2B+arg%28w%29%29%2F2 is when the angle between vectors z and w is less than pi :  |arg(z)-arg(w)| <= pi.

    The second case  arg(z+w) = %28arg%28z%29+%2B+arg%28w%29%29%2F2++%2B+pi is when the angle between vectors z and w is greater than pi :  |arg(z)-arg(w)| > pi.

    Notice that by the modulo of pi,  arg(z+w) = %28arg%28z%29+%2B+arg%28w%29%29%2F2   always.



    Further, the product zw is the unit vector, again, so the same formulas are applicable to vectors zw and 1 = (1,0).

    Notice that arg(zw) = arg(z) + arg(w), so arg(zw+1) is EITHER %28arg%28z%29+%2B+arg%28w%29%29%2F2,  or  %28arg%28z%29+%2B+arg%28w%29%29%2F2+%2B+pi, depending
    on the angle between vectors zw and 1 = (1,0).

    But in any case,  the vectors (z+w) and (zw+1) are EITHER parallel OR anti-parallel (opposite).

    By the modulo of pi,  arg(zw+1) = %28arg%28z%29+%2B+arg%28w%29%29%2F2   always.



    By the rule of argument of quotient for complex numbers, it means that the ratio  %28z%2Bw%29%2F%28zw%2B1%29 is real number.

    This real number is EITHER positive (when the vectors (z+w) and (zw+1) are parallel), 

                          OR   negative (when the vectors (z+w) and (zw+1) are anti-parallel).                     

At this point, the proof is completed.