4u Exam Tomorrow!!! Please Help (1 Viewer)

[coeyz]

Thanks Heaps T______t

 

[youngminii]

You have the answers..
sqrt(5 - 12i) = a + ib
5 - 12i = a^2 - b^2 + 2abi (after you do the working out of (a + ib)^2)
Equating
5 = a^2 - b^2_____(1)
-12 = 2ab
a = -6/b__________(2)
(2) -> (1)
5 = 36/b^2 - b^2
5b^2 = 36 - b^4
b^4 + 5b^2 - 36 = 0
(b^2 + 9)(b^2 - 4) = 0
But, b must be real
b = +-3
Sub into (2)
a = -+2
Therefore, sqrt(5 - 12i) = +-(2 - 3i)

So, |z^2 - 5 + 12i| = |z - 3 + 2i|
Let z = x + yi
|x^2 - y^2 - 5 + (2xy + 12)i| = |x - 3 + (y + 2)i|
(x^2 - y^2 - 5)^2 + (2xy + 12)^2 = (x - 3)^2 + (y + 2)^2
x^4 - x^2y^2 - 5x^2 - x^2y^2 + y^4 + 5y^2 - 5x^2 + 5y^2 + 25 + 4x^2y^2 + 48xy + 144 = x^2 - 6x + 9 + y^2 + 4y + 4
x^4

I have a feeling I did something wrong

Edit: F*CK
Disregard working above
Okay so |z^2 - 5 + 12i| = |z - 3 + 2i|
ie. |z^2 -(5 - 12i)| = |z -(3 - 2i)|
As you should know, |z -(x + iy)| is a circle of centre x, y (with radius being the root of the thing it equals)
So I'm assuming you have to change the first thing to resemble a circle?
so |z^2 -(5 - 12i)| has to become the circle equation
ie. |z -sqrt(5 - 12i)|
Which is |z -(2 - 3i)| by the first part
|z -(2 - 3i)| = |z -(3 - 2i)|
Okay I must've made a mistake somewhere, brb dinner

 

[tommykins]

1. let z = sqrt 5-12i = x+iy
z^2 = x^2-y^2 + 2ixy = 5 - 12i
(1) x^2-y^2 = 5
(2) x^2+y^2 = 13
(3) xy = -6

(1)+(2)
2x^2 = 18
x^2 = 9
x = +-3

sub that into (3) to get your answers.


i'm capped and 2. doesn't seem to load..

 

[coeyz]

i got it now!!!
thankyou all guys