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

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       Log On


   

Question 1171585: 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.
Can you please explain in detail? I'm super confused, can someone please help? Thank you so much.
Answer by math_tutor2020(3817) About Me  (Show Source):
You can put this solution on YOUR website!

Part (a)

Let
z = a+bi
w = c+di
where a,b,c,d are real numbers and i = sqrt(-1) or i^2 = -1

Since |z| = 1, this means
|z| = sqrt(a^2+b^2)
1 = sqrt(a^2+b^2)
1^2 = (sqrt(a^2+b^2))^2
1 = a^2 + b^2
a^2 + b^2 = 1

Through similar algebraic steps, we can say,
|w| = 1
leads to
c^2 + d^2 = 1

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Using our definition of z, let's find 1/z
z+=+a%2Bbi

1%2Fz+=+1%2F%28a%2Bbi%29

1%2Fz+=+%281%2F%28a%2Bbi%29%29%2A%28%28a-bi%29%2F%28a-bi%29%29 This step is done to turn the denominator into a real number.

1%2Fz+=+%281%2A%28a-bi%29%29%2F%28%28a%2Bbi%29%2A%28a-bi%29%29

1%2Fz+=+%28a-bi%29%2F%28a%5E2-%28bi%29%5E2%29 Difference of squares rule

1%2Fz+=+%28a-bi%29%2F%28a%5E2-b%5E2i%5E2%29

1%2Fz+=+%28a-bi%29%2F%28a%5E2-b%5E2%28-1%29%29 Plug in i^2 = -1

1%2Fz+=+%28a-bi%29%2F%28a%5E2%2Bb%5E2%29

1%2Fz+=+%28a-bi%29%2F%281%29 Plug in a^2+b^2 = 1

1%2Fz+=+a-bi



This proves that 1%2Fz is equal to the complex conjugate of z

The steps to proving 1%2Fw is equal to the conjugate of w are effectively the same as shown above. I'll let you do this part.

We'll use the fact that along with in part (b).

==========================================================================

Part (b)

Let
x = complex conjugate of z = a-bi
y = complex conjugate of w = c-di
I'm using x and y instead of the overbar notation because I think the overbar notation is a bit clunky, especially when mixed with fraction bars.

So,
z+x = (a+bi)+(a-bi) = 2a
w+y = (c+di)+(c-di) = 2c
both are real values.


Furthermore,
z*x = 1
w*y = 1
was proven earlier in part (a), just with different notation.
This indicates that zx*wy = 1*1 = 1.


Now if
z = a+bi
w = c+di
Then,
z*w = (a+bi)*(c+di)
z*w = a*(c+di)+bi*(c+di)
z*w = ac+adi+bci+bdi^2
z*w = ac+adi+bci+bd(-1)
z*w = ac+adi+bci-bd
z*w = (ac-bd)+(adi+bci)
z*w = (ac-bd)+(ad+bc)i

Through very similar steps we can say,
x = a-bi
y = c-di
x*y = (ac-bd)-(ad+bc)i

As you can see:
zw+xy = [(ac-bd)+(ad+bc)i] + [(ac-bd)-(ad+bc)i]
zw+xy = [(ac-bd)+(ac-bd)]+[(ad+bc)i-(ad+bc)i]
zw+xy = [2(ac-bd)]+[0(ad+bc)i]
zw+xy = 2(ac-bd)+0i
zw+xy = 2(ac-bd)
which is a real result

To summarize everything so far for part (b), we can add any complex number to its conjugate to get a real result. Similarly, we can multiply any complex number with its conjugate to get a real result. Lastly, zw+xy sorta involves both concepts going on at once which means zw+xy is also a real number.


-------------------------

We build up those statements to be able to say the following

%28z%2Bw%29%2F%28zw%2B1%29

%28%28z%2Bw%29%28xy%2B1%29%29%2F%28%28zw%2B1%29%2A%28xy%2B1%29%29 Multiply numerator and denominator by xy+1

%28%28z%2Bw%29%28xy%2B1%29%29%2F%28%28zw%2Axy%29%2Bzw%2Bxy%2B1%29 Expand out the denominator

%28%28z%2Bw%29%28xy%2B1%29%29%2F%28%28zx%2Awy%29%2B%28zw%2Bxy%29%2B1%29 Group up the terms like such

%28%28z%2Bw%29%28xy%2B1%29%29%2F%281%2B%28zw%2Bxy%29%2B1%29 Plug in zx*wy = 1

%28%28z%2Bw%29%28xy%2B1%29%29%2F%281%2B2%28ac-bd%29%2B1%29 Plug in zw+xy = 2(ac-bd)

%28%28z%2Bw%29%28xy%2B1%29%29%2F%282%28ac-bd%29%2B2%29 Simplify

The denominator 2(ac-bd)+2 is some real number. Let's expand out the numerator and see what we get

%28%28z%2Bw%29%28xy%2B1%29%29%2F%282%28ac-bd%29%2B2%29

%28zxy%2Bz%2Bwxy%2Bw%29%2F%282%28ac-bd%29%2B2%29 FOIL the numerator

%28zx%2Ay%2Bz%2Bwy%2Ax%2Bw%29%2F%282%28ac-bd%29%2B2%29

%281y%2Bz%2B1x%2Bw%29%2F%282%28ac-bd%29%2B2%29 Plug in zx = 1 and wy = 1.

%28y%2Bz%2Bx%2Bw%29%2F%282%28ac-bd%29%2B2%29

%28%28z%2Bx%29%2B%28w%2By%29%29%2F%282%28ac-bd%29%2B2%29 Group up the conjugate pairs

%282a%2B2c%29%2F%282%28ac-bd%29%2B2%29 Plug in z+x = 2a and w+y = 2c

The numerator 2a+2c is a real number.

We could divide every term by 2 to simplify further, but at this point we're effectively done with the proof.

Both the numerator and denominator are real values, so overall %28z%2Bw%29%2F%28zw%2B1%29 is a real number.

Note: The condition zw+%3C%3E+-1 is to prevent the denominator zw%2B1 from being zero.

Further Reading:
https://math.stackexchange.com/questions/427663/prove-if-z-w-1-and-1zw-neq-0-then-zw-over-1zw-is-a-real