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VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Specialist Mathematics => Topic started by: Bestie on April 26, 2014, 05:24:29 pm

Title: complex
Post by: Bestie on April 26, 2014, 05:24:29 pm: if z conjugate is 1/(2+3i), wouldn't z just be 1/(2-3i) cause its a reflection in the x axis?
Title: Re: complex
Post by: kinslayer on April 26, 2014, 05:54:13 pm: Quote from: Bestie on April 26, 2014, 05:24:29 pm
if z conjugate is 1/(2+3i), wouldn't z just be 1/(2-3i) cause its a reflection in the x axis?

The conjugate is 1/(2-3i), yes, but in order to put the number 1/(2+3i) on an Argand diagram (to take the reflection on the x-axis) you would need to express z as z = Re(z) + i*Im(z) first.

If you write z = a + bi then take the conjugate you can see why the conjugate works the way you wrote above.
Title: Re: complex
Post by: Bestie on April 26, 2014, 06:26:27 pm: Oh ok.
Thank you so much

How do I use the completing the square method to factorise
z^2 -(3-2i)z -(4+3i)

So I started it by going
(z-(3-2i)/2)^2 -((3-2i)^2/4) -4-3i
but as I went on it didn't work out.
The ans says (z-(3+sqrt21)/2 +i) and (z-(3-sqrt21)/2 +i)

Thank you for helping
Title: Re: complex
Post by: kinslayer on April 26, 2014, 07:16:11 pm: Quote from: Bestie on April 26, 2014, 06:26:27 pm
How do I use the completing the square method to factorise
z^2 -(3-2i)z -(4+3i)

$z^2 - (3-2i)z - (4 + 3i) = 0$

$(z - 1.5 + i)^2 = \frac{(3 - 2i)^2}{4} + (4 + 3i)$

$(z - 1.5 + i)^2 = \frac{9 -12i -4 + 16 +12i}{4}$

$(z - 1.5 + i)^2 = \frac{21}{4}$

From here you can factorise using $a^2 - b^2 = (a+b)(a-b)$ or you could solve for z and get the factors that way. I solved for z:

$z = \frac{3 - 2i \pm \sqrt{21}}{2}$
Title: Re: complex
Post by: Bestie on April 26, 2014, 07:27:08 pm: I'm sorry I'm still a little lost how did you get to the second step?
Quote from: kinslayer on April 26, 2014, 07:16:11 pm
$z^2 - (3-2i)z - (4 + 3i) = 0$

$(z - 1.5 + i)^2 = \frac{(3 - 2i)^2}{4} + (4 + 3i)$

$(z - 1.5 + i)^2 = \frac{9 -12i -4 + 16 +12i}{4}$

$(z - 1.5 + i)^2 = \frac{21}{4}$

From here you can factorise using $a^2 - b^2 = (a+b)(a-b)$ or you could solve for z and get the factors that way. I solved for z:

$z = \frac{3 - 2i \pm \sqrt{21}}{2}$
Title: Re: complex
Post by: kinslayer on April 26, 2014, 07:51:28 pm: Quote from: Bestie on April 26, 2014, 07:27:08 pm
I'm sorry I'm still a little lost how did you get to the second step?

$(z +b)^2 = z^2 + 2bz + b^2$

If we let $2b = -(3-2i)$ then $b = -1.5 + i$

So: $(z - 1.5 + i)^2 = z^2 -(3-2i)z + \frac{(3-2i)^2}{4}$

That almost looks like what we started with but not quite -- our constant term isn't the same. There are a few ways to deal with this, but the one that I use is where we "add 0" like this:

$z^2 - (3 - 2i)z - (4 + 3i) + \frac{(3-2i)^2}{4} - \frac{(3-2i)^2}{4}$

Now we can arrange the terms into the perfect square. I started by setting the expression to 0 (to solve for z) but you don't have to.
Title: Re: complex
Post by: Bestie on April 29, 2014, 04:43:28 pm: thanks.

With the equation: I know how to solve it to get the hyperbola
Iz-4I-Iz+4I = 2
Ix+yi-4I- Ix+yi+4I = 2
Sqrt[(x-4)^2+y^2] – sqrt[(x+4)^2+y^2] = 2
Sqrt[(x-4)^2+y^2] = 2+ sqrt[(x+4)^2+y^2]
(x-4)^2+y^2 = 4+ 4sqrt[(x+4)^2+y^2] + (x+4)^2+y^2
(x-4)^2= 4+ 4sqrt[(x+4)^2+y^2 + (x+4)^2
X^2 – 8x +16= 4+ 4sqrt[(x+4)^2+y^2] + X^2 + 8x +16
– 8x = 4+ 4sqrt[(x+4)^2+y^2] + 8x
.......

the hyperbola is x^2 - (y^2/15) = 1, but i was wondering how do i find the restriction on the domain cause apparently not both sides of the curve are shown...