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**mathster****Member**- Registered: 2015-10-09
- Posts: 15

The value

is a positive real number. What real number is it?

The problem states that there is a shortcut and big calculations are unnecessary and probably part of a wrong solution.

I have found that the complex number without the power of 72 multiplied by its conjugate is the its magnitude squared. So, the magnitude must be a real number. However, I don't know what to do with this information.

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**bob bundy****Administrator**- Registered: 2010-06-20
- Posts: 8,232

hi mathster

What you have discovered about the number and its complex conjugate is generally true for all complex numbers. But this one has what as its magnitude?

Then you can apply de Moivre's theorem. https://en.wikipedia.org/wiki/De_Moivre%27s_formula

LATER EDIT:

Use a calculator to evaluate (in degrees) ACOS((root3 +1)/(2root2))

Now can you prove this result analytically?

Bob

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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**mathster****Member**- Registered: 2015-10-09
- Posts: 15

bob bundy wrote:

hi mathster

What you have discovered about the number and its complex conjugate is generally true for all complex numbers. But this one has what as its magnitude?

Then you can apply de Moivre's theorem. https://en.wikipedia.org/wiki/De_Moivre%27s_formula

LATER EDIT:

Use a calculator to evaluate (in degrees) ACOS((root3 +1)/(2root2))

Now can you prove this result analytically?

Bob

Hi, bob. Thanks, but I don't really understand trigonometry this well. We haven't studied it in class yet, so I know there's an easier solution. Could you please explain it without using the theorem?

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**bob bundy****Administrator**- Registered: 2010-06-20
- Posts: 8,232

hi mathster

I'm not sure I can but a preliminary first.

There's a theorem in complex numbers as follows:

If you change each of two complex numbers to modulus-argument form then

In words: the product of two complex numbers has modulus equal to the product of the moduli, and argument equal to the sum of the arguments.

Are you familiar with modulus-argument form and this theorem?

LATER EDIT: I'll have to re-think this as 'argument' form uses trig. Whoops!!

SLIGHTLY LATER EDIT: OK. I have an alternative method but it'll take a few multiplications.

Start by calling the bracketed number a + bi

Work out (a + bi)^2, then (a + bi)^4 and finally (a + bi)^6

You'll find that last is a simple complex number, let's say C. As 72 = 6 x 12 all you need to do is to raise C to the power 12.

Bob

ps. Anyone else trying to follow this who does know modulus-argument form, will be able to see where this method came from by calculating the argument of a + bi.

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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