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

## A

• ### 2

• Total voters
0

#### Edema

I need a reply as soon as possible

#### Attachments

• 841.9 KB Views: 13

#### idontknow

$$\displaystyle z\overline{z}=(a+bi)(a-bi)=a^2 - b^2 i^2 = a^2 + b^2 = |z|$$.

• Edema

#### Edema

I mean you solve the second question not the first one ......

#### Edema

$$\displaystyle z\overline{z}=(a+bi)(a-bi)=a^2 - b^2 i^2 = a^2 + b^2 = |z|$$.
Please try and solve the second one

#### topsquark

Math Team
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

• idontknow

#### Edema

I don't know any good solutions to it ....

#### idontknow

I forgot the sqrt code.

#### topsquark

Math Team
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
(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?