# complex num to cartesian form

• Jun 1st 2008, 05:26 AM
taurus
complex 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, 05:37 AM
mr fantastic
Quote:

Originally Posted by taurus
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

Expand the denominator so you get a number of the form a + ib. Then multiply numerator and denominator by the complex conjugate a - ib. Expand numertaor and denominator. Simplify.
• Jun 1st 2008, 05:59 AM
taurus
how do u expand the denominator because i get the following which is wrong:

sqrt(3) - i + sqrt(3)i - 1
• Jun 1st 2008, 06:09 AM
mr fantastic
Quote:

Originally Posted by taurus
how do u expand the denominator because i get the following which is wrong:

sqrt(3) - i - sqrt(3)i - 1

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

Note the correction (in red) of a careless error.
• Jun 1st 2008, 06:18 AM
taurus
and so the comples conjugate is:
$= (\sqrt{3} - 1) + i (\sqrt{3} + 1)$

as i just have to change the sign correct?

so i would get:

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

How do i simplify that now?
• Jun 1st 2008, 06:41 AM
Moo
Hello,

Quote:

Originally Posted by taurus
and so the comples conjugate is:
$= (\sqrt{3} - 1) + i (\sqrt{3} + 1)$

as i just have to change the sign correct?

so i would get:

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

How do i simplify that now?

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

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

$\frac{\sqrt{3}+i}{(\sqrt{3} - 1) - i (\sqrt{3} + 1)}=\frac{(\sqrt{3}+i)((\sqrt{3} - 1)+i (\sqrt{3} + 1))}{4}$
• Jun 1st 2008, 07:26 AM
taurus
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, 07:45 AM
Gusbob
The question you gave was $\frac{1-2i}{3+4i} - \frac{2+i}{5i}$

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

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

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