Hi

How do i express a complex number in exact cartesian form?

for example:

(sqrt(3)+i)/((1-i)(sqrt(3)-i))

Whats the process?

thanks

phil

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- Jun 1st 2008, 04:26 AMtauruscomplex num to cartesian form
Hi

How do i express a complex number in exact cartesian form?

for example:

(sqrt(3)+i)/((1-i)(sqrt(3)-i))

Whats the process?

thanks

phil - Jun 1st 2008, 04:37 AMmr fantastic
- Jun 1st 2008, 04:59 AMtaurus
how do u expand the denominator because i get the following which is wrong:

sqrt(3) - i + sqrt(3)i - 1 - Jun 1st 2008, 05:09 AMmr fantastic
- Jun 1st 2008, 05:18 AMtaurus
and so the comples conjugate is:

$\displaystyle = (\sqrt{3} - 1) + i (\sqrt{3} + 1)$

as i just have to change the sign correct?

so i would get:

$\displaystyle (\sqrt{3} + i)$$\displaystyle (\sqrt{3} - 1) + i (\sqrt{3} + 1)$

----------------------------------

$\displaystyle (\sqrt{3} - 1) - i (\sqrt{3} + 1)$

How do i simplify that now? - Jun 1st 2008, 05:41 AMMoo
Hello,

No, you have to multiply the denominator and the numerator by the conjugate ! :)

So in the denominator, you will have $\displaystyle (\sqrt{3}-1)^2+(\sqrt{3}+1)^2=\dots=4$

$\displaystyle \frac{\sqrt{3}+i}{(\sqrt{3} - 1) - i (\sqrt{3} + 1)}=\frac{(\sqrt{3}+i)((\sqrt{3} - 1)+i (\sqrt{3} + 1))}{4}$ - Jun 1st 2008, 06:26 AMtaurus
heres another example i tried but looking at answer its not quite right:

((1-2i)/(3+4i)) - ((2+i)/5i)

so i first did subtraction of the top and bottom and got:

(-1-3i)/(3-1i)

I then performed division and got this:

((-3+3)/(9+1)) + ((-9+1)/(9+1))i

which simplifies to this:

0 + (-8/10)i

=> -4/5i

But that isint correct as the answer is -2/5

what am i doing wrong? - Jun 1st 2008, 06:45 AMGusbob
The question you gave was $\displaystyle \frac{1-2i}{3+4i} - \frac{2+i}{5i}$

This is two fractions with different denominators. If you recall your basic addition and subtraction of fractions, you don't subtract top and bottom. You need to make a common denominator first, and then subtract the top

$\displaystyle \frac{(1-2i)(5i)}{(3+4i)(5i)} - \frac{(2+i)(3+4i)}{(5i)(3+4i)} \Rightarrow \frac{(1-2i)(5i)-(2+i)(3+4i)}{(3+4i)(5i)}$

Expand that and factorise, and you should get the answer.