Calculate the amount lzl of the complex number

would glad if u guys could give me exemple because i cant find what lzl is

Results 1 to 11 of 11

- Sep 18th 2012, 03:26 AM #1

- Joined
- Sep 2012
- From
- Sweden
- Posts
- 250
- Thanks
- 6

- Sep 18th 2012, 03:34 AM #2

- Joined
- Oct 2009
- Posts
- 5,577
- Thanks
- 790

- Sep 18th 2012, 03:34 AM #3

- Sep 18th 2012, 03:45 AM #4

- Joined
- Sep 2012
- From
- Sweden
- Posts
- 250
- Thanks
- 6

- Sep 18th 2012, 04:00 AM #5
## Re: help

No, first begin by simplifying the numerator:

$\displaystyle (9+2i)(7+6i)=63+54i+14i+12i^2=51+68i$

Then do the same for the denominator, then factor out any constants and multiply the simplified ratio by the conjugate of the denominator to get the original ratio in the form a + bi.

- Sep 18th 2012, 04:09 AM #6

- Joined
- Sep 2012
- From
- Sweden
- Posts
- 250
- Thanks
- 6

- Sep 18th 2012, 05:05 AM #7

- Joined
- Apr 2005
- Posts
- 19,714
- Thanks
- 3002

- Sep 18th 2012, 05:30 AM #8

- Joined
- Sep 2012
- From
- Sweden
- Posts
- 250
- Thanks
- 6

- Sep 18th 2012, 06:10 AM #9
## Re: help

Yes, that is correct. So you now have:

$\displaystyle \frac{51+68i}{-63+35i}=\frac{1}{7}\cdot\frac{51+68i}{-9+5i}$

We factor out the 7 from the denominator to help keep the computations simple.

So, multiply the complex fraction by the conjugate of the denominator:

$\displaystyle \frac{1}{7}\cdot\frac{51+68i}{-9+5i}\cdot\frac{-9-5i}{-9-5i}$

This will cause the denominator to be real, and you will then have the complex number in rectangular form.

- Sep 18th 2012, 06:32 AM #10

- Sep 18th 2012, 03:26 PM #11