I need the solutions for this .......

A

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Dec 2015
936
122
Earth
\(\displaystyle z\overline{z}=(a+bi)(a-bi)=a^2 - b^2 i^2 = a^2 + b^2 = |z|\).
 
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Dec 2019
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Abeokuta
I mean you solve the second question not the first one ......
 

topsquark

Math Team
May 2013
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996
The Astral plane
Please try and solve the second one
The reason you don't get the "correct" answer is because the answer listed is not correct. The numerator works out okay, but not the denominator.

Please note that we typically ask you to show any work that you've done or attempts that you have made.

@idontknow: Is that a typo? The answer is \(\displaystyle |z|^2\), not |z|.

-Dan
 
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Dec 2019
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Abeokuta
I don't know any good solutions to it ....
 
Dec 2019
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Abeokuta
Kkkkkkkk please ooooooouuuu try remember
 

topsquark

Math Team
May 2013
2,385
996
The Astral plane
I don't know any good solutions to it ....
There aren't any solutions to it. I just told you the given answer is wrong. You'll have to ask your instructor about it.

If you want to try to work it out anyway, first get a common denominator and add the fractions. You can use the previous exercise to get the numerator. Letting z = a + ib I got the denominator to be \(\displaystyle 1- 2a + |z|^2\).

I have no idea what "Kkkkkkkk please ooooooouuuu try remember " is supposed to mean.

-Dan
 
Last edited:

skipjack

Forum Staff
Dec 2006
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(a) $z\bar{z}= |z|^2$
(b) The expression given in the original attachment is correct and easy to obtain.

Can you post what you tried (typed, if possible), so that we can identify any slips you made?