What is the polar form of 1/z and why does the sign of the argument change?

[cmcc3119]

Hi There.

I was given this question and the answer:

Find the polar forms of $$1/z$$ where z = $$\sqrt{}3$$ + i


and $$1/z$$ where z = 4$$\sqrt{}3$$ -4i

Answers respectively are:

$$1/2$$ cis($$-\pi/6$$)

$$1/8$$ cis($$\pi/6$$)


Can someone please explain to me why it is that the sign of the argument changes in the answer from + to -, for the first question, and - to +, for the second question?

 

[Dick]

Because the argument of a+bi and the argument of a-bi are negatives of each other? Why? Of course they don't have to be. You could equally well have written the first one as (1/2)*cis(11*pi/6). Why?

 

[cmcc3119]

I understand that that is how the question is written and the sign indicates the point position in the plane. Sorry I worded the question wrong.

Basically, z = $$\sqrt{}3$$ + i

Polar form of this is: 2( cis $$\pi$$/6 )

then working out 1/z the answer is 1/2 (cis -$$\pi$$/6)

I do not know why the sign changes to make $$\pi$$/6 negative and whether you take 1 as r, and z as t, and use the formula for division which is r/t ( cos ($$\pi$$ - $$\phi$$) + isin ($$\pi$$ - $$\phi$$) but when I tried that I still could not come up with a logical explanation it seemed.

 

[Defennder]

$$\frac{1}{\sqrt{3} + i} = \frac{\sqrt{3} -i}{3+1} = \frac{1}{4} (\sqrt{3} - i) = \frac{1}{4} \left [2 \left \angle -\frac{\pi}{6} \right]$$

You could generalise the same thing for any $\mbox{a,b} \in \Re$ for any complex expression a+bi.

EDIT: To expand on this note that by De Moivre's formula:

$$(cos \theta + isin \theta)^n = cos(n \theta) + isin(n \theta)$$

Let $$z = r (cos \theta + isin\theta)$$.
$$z^n = r^n (cos(n \theta) + i sin(n \theta))$$

Let n=-1, as in your case where we want to find the reciprocal.

$$\frac{1}{z} = \frac{1}{r} (cos(-\theta) + isin(-\theta)) = \frac{1}{r} (cos(\theta) - isin(\theta))$$

Thereby the sign changes from '+' to '-'.

 

[Dick]

1/cis(theta)=cis(-theta). Try and show that as an exercise. Hint: show cis(theta)*cis(-theta)=1 because sin(theta)^2+cos(theta)^2=1.

 

[nickbakay]

I have a problem with a question about polar form, could you help me please?

z=3(cos100degree+isin100degree)

give the polar form of 1/z.

 

[Dick]

I have a problem with a question about polar form, could you help me please?

z=3(cos100degree+isin100degree)

give the polar form of 1/z.

Posting your own thread is better than tagging onto another one, but ok. What's cis(100*degree)*cis(-100*degree)? I'm assuming you know cis(a)=cos(a)+i*sin(a).