exact answers

[avogarbro]

If an answer requires an exact answer does it have to be in mixed fraction form (6 2/3) or improper fraction form (20/3)?

And also are trigonometric substitutions required for this course, because i found a question on it in the tsfx 2006 exam1?

[Ahmad]

Err, 6 2/3 is kinda childish if you ask me, I've never done that, but it could be useful when presenting data in a graph for a presentation or something. Doubt it's required.

Trig subs aren't required unless they tell you use the substitution x = sin. Like a build up sort of thing.

[joechan521]

both fractions are alright, but i usually write 20/3

and you need to know how to do trig sub, but most likely they tell u to let x= sin[theta] or something

[avogarbro]

use the quadratic formula to solve z2+ 3z+(1+ 3i)=0

how do i square root the discriminant?

[Ahmad]

Well upon inspection (due to symmetry) z = -i works. so then the other root is (1+3i)/-i which is a special case of Viete's Theorem

[Collin Li]

I would go for the improper fraction. Mixed fractions are for children, like Ahmad said

2z = -3 +/- root(9 - 4(1+3i))
=> 2z = -3 +/- root(5 - 12i)

Here, you have to 5 - 12i into polar form: 13cis(-arctan(12/5))
=> root(5-12i) = root(13)cis(-1/2arctan(12/5))

Convert to Cartesian form:
cos(2x) = 1 - 2sin^2(x)
=> cos(x) = 1 - 2sin^2(x/2)
=> sin(x/2) = +/-root[1/2*(1 - cos(x))]
=> sin(-1/2arctan(12/5)) = +/-root[1/2*(1-cos(-arctan(12/5))]

Let arctan(12/5) = a
Draw a triangle with angle a, opposite length 12, adjacent length 5 (hence hypotenuse length 13).
=> cos(-a) = cos(a) = 5/13

=> sin(-1/2arctan(12/5)) = +/-root[1/2*(1-5/13)] = +/-root(4/13)
Since -1/2arctan(12/5) is in Q4: sin(-1/2arctan(12/5)) is negative.
=> sin(-1/2arctan(12/5)) = -2root(13) / 13

Use sin^2(x) + cos^2(x) = 1 to find cos(-1/2arctan(12/5)):
cos(x) = +/-root[1 - sin^2(x)]
=> cos(-1/2arctan(12/5)) = +/-root[1 - 4/13] = +/-root(9/13)
Since -1/2arctan(12/5) is in Q4: cos(-1/2arctan(12/5)) is positive.
=> cos(-1/2arctan(12/5)) = 3root(13) / 13

=> root(5-12i) = root(13)cis(-1/2arctan(12/5))
= root(13)[ 3root(13)/13 - 2iroot(13) / 13 ]
= 3 - 2i

=> 2z = -3 +/- (3 - 2i)
=> z = -i and z = -3+i

An alternative way of doing it would be:
(5-12i) = z^2, where z = x+yi
=> (x+yi)^2 = x^2 + 2xyi - y^2 = (5-12i)
Equating real and imaginary parts:
5 = x^2 - y^2
-12 = 2xy

This way is much easier, but only in hindsight.

[enwiabe]

... That was part c. of a build-up question. You were supposed to use part b. which gave you z^2 = -1 + 4i (I think)

and had you find 'z', and then you were supposed to use it for the following question. (I just did that question on the exam you just did)

[Collin Li]

... That was part c. of a build-up question. You were supposed to use part b. which gave you z^2 = -1 + 4i (I think)

and had you find 'z', and then you were supposed to use it for the following question. (I just did that question on the exam you just did)

Yeah, I feared that was the case. They can still do something horrible like that on the short answer and give it 4 marks or so. The efficient method would be using Cartesian expansion (x+yi)^2, to solve the root, but most people would stick to de Moivre's theorem methinks.

[avogarbro]

use the quadratic formula to solve z2+ 3z+(1+ 3i)=0

how do i square root the discriminant?

sry for making you do all that typing but the question was z^2+ 3z+(1+ 3i)=0

The question is from the 2006 kilbaha exam 1 Q7c . I just realised you need to use the previous answer to find the the square root of the discriminate. But  thanks for the detailed response from coblin.

[Ahmad]

My way is still the quickest  :wink:

But I don't know if "inspection" is a valid method for the first root. What do you guys reckon?

[Collin Li]

use the quadratic formula to solve z2+ 3z+(1+ 3i)=0

how do i square root the discriminant?

sry for making you do all that typing but the question was z^2+ 3z+(1+ 3i)=0

The question is from the 2006 kilbaha exam 1 Q7c . I just realised you need to use the previous answer to find the the square root of the discriminate. But  thanks for the detailed response from coblin.

Yeah, I was solving that equation. It just looks weird because I wrote it as "2z = ..." I didn't want to put the fraction on the RHS, so I brought the 2 from ".../2" of the quadratic formula to the LHS.

I still solved your equation correctly though.

[Ahmad]

For interest's sake, you can also use a geometrical argument, and the Angle Bisector Theorem to find rt[5 - 12i]. ^^

[joechan521]

use the quadratic formula to solve z2+ 3z+(1+ 3i)=0

how do i square root the discriminant?

sry for making you do all that typing but the question was z^2+ 3z+(1+ 3i)=0

The question is from the 2006 kilbaha exam 1 Q7c . I just realised you need to use the previous answer to find the the square root of the discriminate. But  thanks for the detailed response from coblin.

actually, the 2006 kilbaha exam1 q7c is root 3 not 3

[joechan521]

use the quadratic formula to solve z2+ 3z+(1+ 3i)=0

how do i square root the discriminant?

use quadratic formula u let
a=1, b=3, c=1+3i
discriminent = 9-4-12i = can be written as 9-12i+4i^2  
now thats a perfect square of a^2-2ab+b^2  where a=3, b=2i
so discriminent becomes (3-2i)^2 which cancels your root

then u can find z