Given that and , where z is not 6 . If w is totally imaginary , show that

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- Dec 10th 2008, 04:33 AMmathaddictcomplex number
Given that and , where z is not 6 . If w is totally imaginary , show that

- Dec 10th 2008, 06:47 AMGrandadComplex numbers
Hello again mathaddict.

Quote:

Given that http://www.mathhelpforum.com/math-he...bed5445e-1.gif and http://www.mathhelpforum.com/math-he...866f0d3b-1.gif, where z is not 6 . If w is totally imaginary , show that http://www.mathhelpforum.com/math-he...ebd063ba-1.gif

I think the expression should be .

You then:

- Replace by in the top and bottom
- Collect together reals and imaginaries on the top and on the bottom
- Multiply top and bottom by , which is the conjugate of the bottom

**Don't**multiply everything out at this stage. Just find the**real**part of the top. This must be zero if*w*is wholly imaginary. This will then give you

Grandad - Dec 10th 2008, 07:07 AMSoroban
Hello, mathaddict!

*Grandad*caught the typo . . .

Quote:

Given: .

If is totally imaginary , show that

Substituting , we have: .

Multiply top and bottom by

. .

. .

. .

Since is purely imaginary,

Therefore: .

- Dec 11th 2008, 02:36 AMmathaddictRe :
Thank you very much again , Soroban and Grandad !!