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
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)$
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$\displaystyle (\sqrt{3} - 1) - i (\sqrt{3} + 1)$
How do i simplify that now?
Hello,
No, you have to multiply the denominator and the numerator by the conjugate !
So in the denominator, you will have $\displaystyle (\sqrt{3}-1)^2+(\sqrt{3}+1)^2=\dots=4$
$\displaystyle \frac{\sqrt{3}+i}{(\sqrt{3} - 1) - i (\sqrt{3} + 1)}=\frac{(\sqrt{3}+i)((\sqrt{3} - 1)+i (\sqrt{3} + 1))}{4}$
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?
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.