F maths AS level help: writing complex numbers in mod-arg form...

OP

Thanks both of you, I rationalised before but got stuck from there. But i've got it now, thanks

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Reply 4

Original post by muf_mur

just rationalise the denominator

You have the right idea, but the wrong terminology: we don't particularly want to rationalise the denominator, but to make it real, by using an application of the d-o-2-s. For example, if the problem was based on the expression

\frac{3}{1+3i}

, then there would be no irrational numbers in sight, but the method would remain the same.

Reply 5

TenOfThem

18

Original post by atsruser

You have the right idea, but the wrong terminology: we don't particularly want to rationalise the denominator, but to make it real, by using an application of the d-o-2-s. For example, if the problem was based on the expression

\frac{3}{1+3i}

, then there would be no irrational numbers in sight, but the method would remain the same.

There may be no irrational numbers

But do you believe that the denominator is rational

After multiplying by the complex conjugate - will the denominator be rational

Reply 6

Original post by TenOfThem

There may be no irrational numbers

But do you believe that the denominator is rational

After multiplying by the complex conjugate - will the denominator be rational

The denominator is not rational due to the presence of an imaginary part. However, if we start with

\frac{3}{3^{1/4}+i}

, then we will have:

\displaystyle \frac{3}{3^{1/4}+i} = \frac{3}{3^{1/4}+i} \times \frac{3^{1/4}-i}{3^{1/4}-i} = \frac{3(3^{1/4}-i)}{\sqrt{3}+1}

whose denominator is still irrational.

So the supposed "rationalisation", in this case, has generated an irrational number in the denominator. It has, however, still generated a real number, which is why it is incorrect to refer to this method as "rationalisation"; we need an expression like "realising", or something.

The point is the presence of rational or irrational numbers in the denom. is irrelevant; we are interested in the presence or absence of real or complex numbers.

(edited 10 years ago)

Reply 7

OP

Thought i had it but I don't. The answer is 3/2(cos(-π/3)+isin(-π/3)).
Someone explain this to me so I can correct my misconceptions.

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Reply 8

tigerz

Original post by emilierichards

Thought i had it but I don't. The answer is 3/2(cos(-π/3)+isin(-π/3)).
Someone explain this to me so I can correct my misconceptions.

Posted from TSR Mobile

Heya, first of all you need times by the complex conjugate:

\frac{3}{1+i\sqrt{3}}=\frac{3}{1+i\sqrt{3}}\times\frac{1-i\sqrt{3}}{1-i\sqrt{3}}....

your answer will look similar to this:

\frac{a-(b\sqrt{3})i}{4}

From this point i'll solve for a similar problem so you can follow the steps
You can then find |r| by using pythagoras and then that will be the r(cosx+isinx) e.g for

\frac{2-(2\sqrt{3})i}{4}

\left | r \right |=\sqrt{(0.5)^2+(-0.5\sqrt{3})^2}= 1

Then you need to find theta, if you draw a quick sketch you'll see that the point is in the fourth quadrant so your answer is negative..

-\frac{2\sqrt{3}}{4} \div \frac{2}{4}= -\sqrt{3}

you will do then do

\tan^{-1}(-\sqrt{3})

so the final answer is...

1(cos(-\frac{\pi}{3})+isin(-\frac{\pi}{3}))

If you want the answer i'm happy to help

(edited 10 years ago)

Reply 9

Original post by emilierichards

Thought i had it but I don't. The answer is 3/2(cos(-π/3)+isin(-π/3)).
Someone explain this to me so I can correct my misconceptions.

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\frac{3}{1+i\sqrt{3}} = \frac{3}{1+i\sqrt{3}} \times \frac{1-i\sqrt{3}}{1-i\sqrt{3}}

What's the next step?

Reply 10

Original post by tigerz

Heya, first of all you need times by the complex conjugate:

\frac{3}{1-\sqrt{3}}=\frac{3}{1-\sqrt{3}}\times\frac{1+\sqrt{3}}{1+\sqrt{3}}....

That isn't the question that was posed.

Reply 11

tigerz

Original post by atsruser

That isn't the question that was posed.

damn >.< I swear I make a ridiculous amount of mistakes, sorry about that, will correct :smile:

edit: switched the signs :facepalm:

should be correct now

(edited 10 years ago)

Reply 12

Original post by tigerz

damn >.< I swear I make a ridiculous amount of mistakes, sorry about that, will correct :smile:

edit: switched the signs :facepalm:

should be correct now

You'd better put in the

i

, too

Reply 13

TenOfThem

18

Original post by atsruser

The denominator is not rational due to the presence of an imaginary part. However, if we start with

\frac{3}{3^{1/4}+i}

, then we will have:

\displaystyle \frac{3}{3^{1/4}+i} = \frac{3}{3^{1/4}+i} \times \frac{3^{1/4}-i}{3^{1/4}-i} = \frac{3(3^{1/4}-i)}{\sqrt{3}+1}

whose denominator is still irrational.

So the supposed "rationalisation", in this case, has generated an irrational number in the denominator. It has, however, still generated a real number, which is why it is incorrect to refer to this method as "rationalisation"; we need an expression like "realising", or something.

The point is the presence of rational or irrational numbers in the denom. is irrelevant; we are interested in the presence or absence of real or complex numbers.

My mistake - I missed the root3 :smile:

I just felt that the correction was un-necessary even in terms of terminology

Reply 14

tigerz

Original post by atsruser

You'd better put in the

i

, too

Right, I think I need to go back to bed :colondollar:

Thank you haha

Reply 15

Original post by TenOfThem

My mistake - I missed the root3 :smile:

You didn't miss the root three - that was only in my example. I was merely pointing out an example that shows that we can perfectly well "realise" a complex denominator without having to rationalise it.

Making real and making rational are independent operations.

I just felt that the correction was un-necessary even in terms of terminology

This terminology is indeed used, but it's clearly misleading, and wrong, and it grates on me. But it's unnecessary in the sense that the OP is probably more confused than she was at the start.

Reply 16

OP

Thank you so much.

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Reply 17

Original post by emilierichards

Thank you so much.

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So did you figure it out then? I though that you were confused?

Reply 18