# z times z

• Nov 8th 2008, 09:24 PM
magentarita
z 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 PM
Jhevon
Quote:

Originally Posted by magentarita
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 =

what are you having trouble with?

you were given the rule

here you have a = 3 and b = -4

• Nov 9th 2008, 05:20 AM
magentarita
but....
Quote:

Originally Posted by Jhevon
what are you having trouble with?

you were given the rule

here you have a = 3 and b = -4

The whole question is: Given z = 3 - 4i, write the question in the form
a + bi.

For z times z, the answer is 5(x + 3) (x - 2)^2 (x + 1).

How do I get this answer?
• Nov 9th 2008, 10:01 AM
Jhevon
Quote:

Originally Posted by magentarita
The whole question is: Given z = 3 - 4i, write the question in the form
a + bi.

For z times z, the answer is 5(x + 3) (x - 2)^2 (x + 1).

How do I get this answer?

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 AM
HallsofIvy
Quote:

Originally Posted by magentarita
The whole question is: Given z = 3 - 4i, write the question in the form
a + bi.

For z times z, the answer is 5(x + 3) (x - 2)^2 (x + 1).

How do I get this answer?

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 PM
magentarita
the x letters
Quote:

Originally Posted by Jhevon
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?!

I think the x letters could be a typo in the math book. Math books are famous for having tons of wrong answers to great questions.
• Nov 11th 2008, 06:58 AM
Jhevon
Quote:

Originally Posted by magentarita
I think the x letters could be a typo in the math book. Math books are famous for having tons of wrong answers to great questions.

not all math books, but some do. this is definitely a typo (are you sure you're looking in the right section?) there is no where for an x to appear.
• Nov 11th 2008, 09:01 PM
magentarita
it was a..............
Quote:

Originally Posted by HallsofIvy
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.

It was a textbook error.
• Nov 11th 2008, 09:02 PM
magentarita
great work.......
Quote:

Originally Posted by Jhevon
what are you having trouble with?

you were given the rule

here you have a = 3 and b = -4

Fantastic!
• Nov 11th 2008, 09:03 PM
magentarita
I see..............
Quote:

Originally Posted by Jhevon
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?!

I got it now. Thank you.
• Nov 11th 2008, 09:04 PM
magentarita
yes.........
Quote:

Originally Posted by Jhevon
not all math books, but some do. this is definitely a typo (are you sure you're looking in the right section?) there is no where for an x to appear.

Yes, I looked at the section several times before deciding to place the question in this forum.