Question 1171585
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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+bi}}}


{{{1/z = 1/(a+bi)}}}


{{{1/z = (1/(a+bi))*((a-bi)/(a-bi))}}} This step is done to turn the denominator into a real number.


{{{1/z = (1*(a-bi))/((a+bi)*(a-bi))}}}


{{{1/z = (a-bi)/(a^2-(bi)^2)}}} Difference of squares rule


{{{1/z = (a-bi)/(a^2-b^2i^2)}}}


{{{1/z = (a-bi)/(a^2-b^2(-1))}}} Plug in i^2 = -1


{{{1/z = (a-bi)/(a^2+b^2)}}}


{{{1/z = (a-bi)/(1)}}} Plug in a^2+b^2 = 1


{{{1/z = a-bi}}}


*[Tex \Large \frac{1}{z} = \overline{z}]


This proves that {{{1/z}}} is equal to the complex conjugate of {{{z}}}


The steps to proving {{{1/w}}} 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 *[Tex \large z*\frac{1}{z}=z*\overline{z}=1] along with *[Tex \large w*\frac{1}{w}=w*\overline{w}=1] in part (b).


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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.



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


{{{(z+w)/(zw+1)}}}


{{{((z+w)(xy+1))/((zw+1)*(xy+1))}}} Multiply numerator and denominator by xy+1


{{{((z+w)(xy+1))/((zw*xy)+zw+xy+1)}}} Expand out the denominator


{{{((z+w)(xy+1))/((zx*wy)+(zw+xy)+1)}}} Group up the terms like such


{{{((z+w)(xy+1))/(1+(zw+xy)+1)}}} Plug in zx*wy = 1


{{{((z+w)(xy+1))/(1+2(ac-bd)+1)}}} Plug in zw+xy = 2(ac-bd)


{{{((z+w)(xy+1))/(2(ac-bd)+2)}}} Simplify


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


{{{((z+w)(xy+1))/(2(ac-bd)+2)}}}


{{{(zxy+z+wxy+w)/(2(ac-bd)+2)}}} FOIL the numerator


{{{(zx*y+z+wy*x+w)/(2(ac-bd)+2)}}}


{{{(1y+z+1x+w)/(2(ac-bd)+2)}}} Plug in zx = 1 and wy = 1. 


{{{(y+z+x+w)/(2(ac-bd)+2)}}}


{{{((z+x)+(w+y))/(2(ac-bd)+2)}}} Group up the conjugate pairs


{{{(2a+2c)/(2(ac-bd)+2)}}} 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 {{{(z+w)/(zw+1)}}} is a real number.


Note: The condition {{{zw <> -1}}} is to prevent the denominator {{{zw+1}}} from being zero. 


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