z times z

**magentarita** · Nov 8th 2008, 10:24 PM

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 =

**Jhevon** · Nov 8th 2008, 10:39 PM

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

just follow the rule

**magentarita** · Nov 9th 2008, 06:20 AM

Originally Posted by Jhevon

what are you having trouble with?

you were given the rule

here you have a = 3 and b = -4

just follow the rule

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?

**Jhevon** · Nov 9th 2008, 11:01 AM

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 $z = a + ib$

then

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

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

where are your x's coming from?

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

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

still, where are you getting your x's from?!

**HallsofIvy** · Nov 9th 2008, 11:09 AM

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 $z\overline{z}= 3^2+ (-4)^2$ , a positive real number, not some product of algebraic expressions. Perhaps you accidently looked up the answer to the wrong question.

**magentarita** · Nov 9th 2008, 10:18 PM

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 $z = a + ib$

then

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

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

where are your x's coming from?

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

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

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.

**Jhevon** · Nov 11th 2008, 07:58 AM

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.

**magentarita** · Nov 11th 2008, 10:01 PM

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 $z\overline{z}= 3^2+ (-4)^2$ , 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.

**magentarita** · Nov 11th 2008, 10:02 PM

Originally Posted by Jhevon

what are you having trouble with?

you were given the rule

here you have a = 3 and b = -4

just follow the rule

Fantastic!

**magentarita** · Nov 11th 2008, 10:03 PM

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 $z = a + ib$

then

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

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

where are your x's coming from?

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

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

still, where are you getting your x's from?!

I got it now. Thank you.

**magentarita** · Nov 11th 2008, 10:04 PM

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.

Thread: z times z

LinkBack

Thread Tools

Display

z times z

but....

the x letters

it was a..............

great work.......

I see..............

yes.........

Similar Math Help Forum Discussions

[SOLVED] |a\times b| = LCM[|a|, |b|]

3 times differentiable, but not 3 times continuously differentiable?

I have tried this problem 8 times, I need some help!

Stopping times

any permutation times a k-cycle times the inverse is a k-cycle

Search Tags