SOLUTION: Use z = a + bi and w = c + di to show that (z • w) bar = z bar • w bar. Note: one bar line over (z • w).

Algebra ->  Complex Numbers Imaginary Numbers Solvers and Lesson  -> Lesson -> SOLUTION: Use z = a + bi and w = c + di to show that (z • w) bar = z bar • w bar. Note: one bar line over (z • w).       Log On


   



Question 1207460: Use z = a + bi and w = c + di to show that (z • w) bar = z bar • w bar.

Note: one bar line over (z • w).

Found 2 solutions by ikleyn, math_tutor2020:
Answer by ikleyn(52787) About Me  (Show Source):
You can put this solution on YOUR website!
.

Step 1.  Multiply (a+bi) by (c+di);  then take the bar over this product.

         it will be (z* w) bar.



Step 2.  Take bars first, transforming  a+bi  to  a-bi
                          and           c+di  to  c-di.

         Now multiply  (a-bi)  by  (c-di).

         It will be z bar * w bar.



Step 3.  Now compare your results in steps 1 and 2.

         If you calculated everything correct, you must get identical results,
         which will prove your statement.

         It you will get different results,  then check and re-check, double check and cross-check,
         until you find and correct all possible mistakes.


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On complex numbers,  see introductory lessons
    - Complex numbers and arithmetical operations on them
    - Complex plane
    - Addition and subtraction of complex numbers in complex plane
    - Multiplication and division of complex numbers in complex plane
in this site.



Answer by math_tutor2020(3817) About Me  (Show Source):
You can put this solution on YOUR website!

a,b,c,d are real numbers
i = sqrt(-1) represents an imaginary number

z = a+bi
zBar = horizontal bar over "a+bi" = complex conjugate of z
zBar = a-bi
w = c+di
wBar = c-di

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-bd) + (ad+bc)i
(z*w)bar = (ac-bd) - (ad+bc)i
We'll return to this later.

zBar*wBar = (a-bi)*(c-di)
zBar*wBar = a*(c-di)-bi*(c-di)
zBar*wBar = ac-adi-bci+bdi^2
zBar*wBar = ac-adi-bci+bd*(-1)
zBar*wBar = ac-adi-bci-bd
zBar*wBar = (ac-bd)+(-adi-bci)
zBar*wBar = (ac-bd)-(ad+bc)i
This is an identical match with the conclusion of the previous paragraph.

Therefore we have shown that (z*w)bar = zBar*wBar is indeed the case.