NOTE: Thezin red below has a bar line over it.

Use the rule z times z = a^2 + b^2 to solve the question below when z = 3 - 4i and w = 8 + 3i.

Solve z timesz =

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- Nov 8th 2008, 09:24 PMmagentaritaz times z
NOTE: The

**z**in red below has a bar line over it.

Use the rule z times z = a^2 + b^2 to solve the question below when z = 3 - 4i and w = 8 + 3i.

Solve z times**z =** - Nov 8th 2008, 09:39 PMJhevon
- Nov 9th 2008, 05:20 AMmagentaritabut....
- Nov 9th 2008, 10:01 AMJhevon
ok, i have no idea where you are getting (x + 3) and (x - 2) from. are you leaving out a part of the question?

the rule is: if $\displaystyle z = a + ib$

then

(1) $\displaystyle z^2 = (a + ib)(a + ib) = a^2 - b^2 + 2abi$

(2) $\displaystyle |z|^2 = a^2 + b^2$

where are your x's coming from?

EDIT: Oh! i see now, you put red z to mean $\displaystyle \bar{z}$!

ok, so $\displaystyle z \bar{z} = (a + ib)(a - ib) = a^2 + b^2 = |z|^2$

still, where are you getting your x's from?! - Nov 9th 2008, 10:09 AMHallsofIvy
What you need to do is go back and read the question CAREFULLY.

For one thing it surely does not say "write the QUESTION in the form a+bi". Perhaps it says "write the ANSWER in the form a+ bi.

But then "5(x+ 3)(x-2)^2(x+1)" is NOT in the form a+ bi!

As you said before $\displaystyle z\overline{z}= 3^2+ (-4)^2$, a positive real number, not some product of algebraic expressions. Perhaps you accidently looked up the answer to the wrong question. - Nov 9th 2008, 09:18 PMmagentaritathe x letters
- Nov 11th 2008, 06:58 AMJhevon
- Nov 11th 2008, 09:01 PMmagentaritait was a..............
- Nov 11th 2008, 09:02 PMmagentaritagreat work.......
- Nov 11th 2008, 09:03 PMmagentaritaI see..............
- Nov 11th 2008, 09:04 PMmagentaritayes.........