1. Complex numbers: 2 problems.

Hi, I have two problems dealing with complex numbers, ill show you them and then what I try to do:

Given $\displaystyle z = x + iy, i = \sqrt-1$ So yeah y = the imaginary and z = x(real)

1. $\displaystyle z + 1 - i = (1+6) / (3 - i)$ Answer given: x = -4/5, y =7/5

^ This one I multiplied by the conjugate 3 + i on the right side. I got:

$\displaystyle z + 1 - i = (21 + 7i)/ 10$ and now do I'm unsure what to do.

2. $\displaystyle (2 + \sqrt-2)(1 - \sqrt-8)$

I'm not sure if I know how to multiply this out correctly...

$\displaystyle 2 - 2\sqrt-8 - \sqrt-2 + 2\sqrt4$?

2. Originally Posted by Kaln0s
Hi, I have two problems dealing with complex numbers, ill show you them and then what I try to do:

Given $\displaystyle z = x + iy, i = \sqrt-1$ So yeah y = the imaginary and z = x(real)

1. $\displaystyle z + 1 - i = (1+6) / (3 - i)$ Answer given: x = -4/5, y =7/5

^ This one I multiplied by the conjugate 3 + i on the right side. I got:

$\displaystyle z + 1 - i = (21 + 7i)/ 10$ and now do I'm unsure what to do.

2. $\displaystyle (2 + \sqrt-2)(1 - \sqrt-8)$

I'm not sure if I know how to multiply this out correctly...

$\displaystyle 2 - 2\sqrt-8 - \sqrt-2 + 2\sqrt4$?
For problem 1. it might have been easier to multiply both sides of the equation by (3-i), as opposed to multiplying the RHS by (3+i)/(3+i). I suggest you try that strategy!

2. Just use foil, but remember that $\displaystyle \sqrt{-a} = i\sqrt{a}$, hence your expression should be:

$\displaystyle (2 + i\sqrt{2})(1 - i\sqrt{8}) = 2 -2i\sqrt{8}+i\sqrt{2}-i^2 \sqrt{2}.\sqrt{8}$

Then remember that $\displaystyle i^2 = \sqrt{-1}^2 = -1$ and that $\displaystyle \sqrt{a}.\sqrt{b} = \sqrt{ab}$, hence:

$\displaystyle = 2 -2i\sqrt{8}+i\sqrt{2}+ \sqrt{16}$

$\displaystyle = 2 -2i\sqrt{8}+i\sqrt{2}+ 4$

Now split into real and imaginary parts!

$\displaystyle = (2+4) +i(\sqrt{2}-2\sqrt{8})$

PS: Are you sure you wrote out 1. correctly? Since the numerator of the RHS is just 6+1 = 7.

3. Originally Posted by Kaln0s
Hi, I have two problems dealing with complex numbers, ill show you them and then what I try to do:

Given $\displaystyle z = x + iy, i = \sqrt-1$ So yeah y = the imaginary and z = x(real)

1. $\displaystyle z + 1 - i = (1+6) / (3 - i)$ Answer given: x = -4/5, y =7/5

^ This one I multiplied by the conjugate 3 + i on the right side. I got:

$\displaystyle z + 1 - i = (21 + 7i)/ 10$ and now do I'm unsure what to do.

2. $\displaystyle (2 + \sqrt-2)(1 - \sqrt-8)$

I'm not sure if I know how to multiply this out correctly...

$\displaystyle 2 - 2\sqrt-8 - \sqrt-2 + 2\sqrt4$?

1. recheck the problem ... is the (1+6) in the numerator on the right side correct?

2. $\displaystyle (2 + i\sqrt{2})(1 - 2i\sqrt{2}) =$

$\displaystyle 2 + i\sqrt{2} - 4i\sqrt{2} + 4 =$

$\displaystyle 6 - 3i\sqrt{2}$

4. Originally Posted by skeeter
1. recheck the problem ... is the (1+6) in the numerator on the right side correct?

2. $\displaystyle (2 + i\sqrt{2})(1 - 2i\sqrt{2}) =$

$\displaystyle 2 + i\sqrt{2} - 4i\sqrt{2} + 4 =$

$\displaystyle 6 - 3i\sqrt{2}$
I uploaded both problems, and they both show the answer. The deal is it's a really bad printout, so you tell me if it's 1+6 :P.