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## #1 2008-07-09 11:15:06

Daniel123
Member
Registered: 2007-05-23
Posts: 663

### Reasonably easy, but reasonably nice, complex numbers question.

I suppose I better state the source of the question:

Oxford Further Pure 1 , Mark Rowland.

Last edited by Daniel123 (2008-07-09 11:28:09)

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## #2 2008-07-09 21:55:12

JaneFairfax
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Registered: 2007-02-23
Posts: 6,868

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## #3 2008-07-09 22:42:50

mathsyperson
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Registered: 2005-06-22
Posts: 4,900

### Re: Reasonably easy, but reasonably nice, complex numbers question.

You might need a) to do b), it depends how much of a stickler the marker is about the word 'hence'.

Why did the vector cross the road?
It wanted to be normal.

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## #4 2008-07-10 03:01:41

Daniel123
Member
Registered: 2007-05-23
Posts: 663

### Re: Reasonably easy, but reasonably nice, complex numbers question.

I spotted that too Jane, but if you've already done part a) it isn't exactly difficult to use it for b).

Out of interest Mathsy and Jane, how did you do pat a)? I've seen a few different ways of doing it.

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## #5 2008-07-10 03:08:41

mathsyperson
Moderator
Registered: 2005-06-22
Posts: 4,900

### Re: Reasonably easy, but reasonably nice, complex numbers question.

I expanded the (a+bi)², equated it with c+di and split the real and imaginary parts to get:

a² - b² = c; 2ab = d.

Rearranging the second and substituting into the first gets a² - (d/2a)² = c, which can be made into a quadratic involving a² (ie. l(a²)² + m(a²) + n = 0).

Solving that gives the required result.

Why did the vector cross the road?
It wanted to be normal.

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## #6 2008-07-10 03:10:49

Daniel123
Member
Registered: 2007-05-23
Posts: 663

### Re: Reasonably easy, but reasonably nice, complex numbers question.

Ahh that was one of the methods I saw. I did it differently. I started off the same way as you, but then I also equated moduli, which gave me enough information to get to the result.

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