Complex Numbers in Polar Form

**alrightgeez** · March 8th 2010, 05:43 AM

I'm not sure where to start with this question at all, some help would be appreciated

A point Z in an Argand diagram represents $z = e^i \theta$

By writing z + 1 as $e^(1/2i\theta) (e^(1/2i\theta) + e^(-1/2i\theta)$ , find the modulus and argument of z+1.

Hmm, the editing has gone badly, it should be:

e^(itheta) in the first and
e^(0.5itheta)(e^(0.5itheta) + e^(-0.5itheta)) in the second.

**Opalg** · March 8th 2010, 06:50 AM

Originally Posted by alrightgeez

I'm not sure where to start with this question at all, some help would be appreciated

A point Z in an Argand diagram represents $z = e^i \theta$

By writing z + 1 as $e^{(1/2)i\theta} (e^{(1/2)i\theta} + e^{-(1/2)i\theta})$ , find the modulus and argument of z+1.

LaTeX tip: use braces { } rather than parentheses ( ) to enclose the superscripts.

Write that expression for z + 1 with the factors in the opposite order: $z+1 = (e^{(1/2)i\theta} + e^{-(1/2)i\theta})e^{(1/2)i\theta}$ . Then remember that a complex number has a unique polar form $re^{i\theta}$ with r a positive real number. Notice also that $e^{(1/2)i\theta} + e^{-(1/2)i\theta} = 2\cos\theta$ , which is real (though it is not always positive).

**alrightgeez** · March 8th 2010, 09:38 AM

Originally Posted by Opalg

LaTeX tip: use braces { } rather than parentheses ( ) to enclose the superscripts.

Write that expression for z + 1 with the factors in the opposite order: $z+1 = (e^{(1/2)i\theta} + e^{-(1/2)i\theta})e^{(1/2)i\theta}$ . Then remember that a complex number has a unique polar form $re^{i\theta}$ with r a positive real number. Notice also that $e^{(1/2)i\theta} + e^{-(1/2)i\theta} = 2\cos\theta$ , which is real (though it is not always positive).

Thanks so much for that, I managed to work it out by myself in my head as I was out for a jog but thanks anyway.

The next bit has me stuck though as it doesn't seem to cancel down correctly to the given answer.

I'm told to use a similar method for 'z-1' (meaning $e^{i\theta} - 1$ )

So I write it out in the same way as: $e^{1/2\theta i}(e^{1/2\theta i} - e^{-1/2\theta i})$

This is my working:

$e^{1/2\theta i}(e^{1/2\theta i} - e^{-1/2\theta i})$
= $e^{1/2\theta i}(cos(1/2\theta) + isin(1/2\theta) - cos(-1/2\theta) - isin(-1/2\theta)))$
= $e^{1/2\theta i}(cos(1/2\theta) + isin(1/2\theta) - cos(1/2\theta) + isin(1/2\theta)))$
= $e^{1/2\theta i}(2isin(1/2\theta))$
= $2isin(1/2\theta)cos(1/2\theta) - 2sin^2(1/2\theta)$

Then try to find the modulus by squaring both terms:

$4sin^4(1/2\theta) - 4sin^2(1/2\theta)cos^2(1/2\theta)$
$4sin^4(1/2\theta) - 4sin^2(1/2\theta)(1 - sin^2(1/2\theta))$
$8sin^4(1/2\theta) - 4sin^2(1/2\theta)$

However the answer given in the book is $2sin(1/2\theta)$ , which the terms above doesn't cancel to. Can anyone see where I've gone wrong?

**Opalg** · March 8th 2010, 12:20 PM

Originally Posted by alrightgeez

Thanks so much for that, I managed to work it out by myself in my head as I was out for a jog but thanks anyway.

The next bit has me stuck though as it doesn't seem to cancel down correctly to the given answer.

I'm told to use a similar method for 'z-1' (meaning $e^{i\theta} - 1$ )

So I write it out in the same way as: $e^{1/2\theta i}(e^{1/2\theta i} - e^{-1/2\theta i})$

This is my working:

$e^{1/2\theta i}(e^{1/2\theta i} - e^{-1/2\theta i})$
= $e^{1/2\theta i}(cos(1/2\theta) + isin(1/2\theta) - cos(-1/2\theta) - isin(-1/2\theta)))$
= $e^{1/2\theta i}(cos(1/2\theta) + isin(1/2\theta) - cos(1/2\theta) + isin(1/2\theta)))$
= $e^{1/2\theta i}(2isin(1/2\theta))$ Stop right there! You're almost finished at this point.
= $2isin(1/2\theta)cos(1/2\theta) - 2sin^2(1/2\theta)$

Then try to find the modulus by squaring both terms:

$4sin^4(1/2\theta) - 4sin^2(1/2\theta)cos^2(1/2\theta)$
$4sin^4(1/2\theta) - 4sin^2(1/2\theta)(1 - sin^2(1/2\theta))$
$8sin^4(1/2\theta) - 4sin^2(1/2\theta)$

However the answer given in the book is $2sin(1/2\theta)$ , which the terms above doesn't cancel to. Can anyone see where I've gone wrong?

The next line after "Stop right there!" should be
$= (2\sin((1/2)\theta))ie^{(1/2)\theta i}$ . Notice that $2\sin((1/2)\theta)$ is real, and $ie^{(1/2)\theta i} = e^{(1/2)\pi i}e^{(1/2)\theta i} = e^{(1/2)(\pi+\theta)i}$ . So the modulus-argument form of z–1 is $2\sin((1/2)\theta)e^{(1/2)(\pi+\theta)i}$ , from which you can read off the modulus as $2\sin((1/2)\theta)$ (provided that $\theta$ is chosen so that $2\sin((1/2)\theta)$ is positive).

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