• Dec 26th 2010, 03:36 PM
aonin
hey guys
1. I proved that z^-1=conj z, converting into mod-arg form and by doing de moivres theorem so this must therefore be true
1/(2+3i)=2-3i

2. yet when i substituted z for an actual complex number eg. z=2+3i
and then did the realisation the answer is different
1/(2+3i) X (2-3i)/(2-3i)=(2-3i)/13

3. and i learned of a complex number rule where
z^-1=conj z/(mod z)^2
and this rule is consistent with what got ... which was (2-3i)/13

so now the question finally is why are there 2 answers that seem right?
• Dec 26th 2010, 03:41 PM
Drexel28
Quote:

Originally Posted by aonin
hey guys
1. I proved that z^-1=conj z, converting into mod-arg form and by doing de moivres theorem so this must therefore be true
1/(2+3i)=2-3i

2. yet when i substituted z for an actual complex number eg. z=2+3i
and then did the realisation the answer is different
1/(2+3i) X (2-3i)/(2-3i)=(2-3i)/13

3. and i learned of a complex number rule where
z^-1=conj z/(mod z)^2
and this rule is consistent with what got ... which was (2-3i)/13

so now the question finally is why are there 2 answers that seem right?

It's definitely untrue that $\displaystyle z^{-1}=\bar{z}$ since as you noted for $\displaystyle z\ne 0$ you have that $\displaystyle z^{-1}=\frac{\bar{z}}{|z|^2}$ from where it quickly follows that $\displaystyle \bar{z}=z^{-1}\implies z\in\mathbb{S}^1$ and the converse can be checked quite readily.
• Dec 26th 2010, 03:42 PM
aonin
but it is true z^-1=conj z through demoivres theorem though?. what is the symbol that looks like an E, sorry im new here :D
• Dec 26th 2010, 03:44 PM
dwsmith
What were you asked to do originally?

You have all these work based on something wrong.

Are you asked to prove or disprove if $\displaystyle z^{-1}=\bar{z} \ \mbox{?}$ If so, an example of why it doesn't suffices which you have. If not, what are you trying to do?
• Dec 26th 2010, 03:46 PM
Plato
There are several mistakes in your OP.
Consider this fact: $\displaystyle z^{ - 1} = \dfrac{1}{z} = \dfrac{{\overline z }}{{\left| z \right|^2 }}$.

What does that do to what you posted?
• Dec 26th 2010, 03:50 PM
Drexel28
Quote:

Originally Posted by aonin
but it is true z^-1=conj z through demoivres theorem though?. what is the symbol that looks like an E, sorry im new here :D

I don't see how DeMoivre's theorem: $\displaystyle \displaystyle (\cos(\theta)+i\sin(\theta))^n=\cos(n\theta)+i\sin (n\theta)$, implies this. Aha! I bet I know where you got messed up! You can get from DeMoivre's theorem that $\displaystyle \left(\cos(\theta)+i\sin(\theta)\right)^{-1}=\cos\left(-1\theta\right)+i\sin(-\theta)=\cos(\theta)-i\sin(\theta)=\overline{\cos(\theta)+i\sin(\theta) }$ and you had a fuzzy idea that every number can be represented as something 'like' $\displaystyle \cos(\theta)+i\sin(\theta)$ but you forgot that it's actually $\displaystyle r(\cos(\theta)+i\sin(\theta))$ where $\displaystyle r=|z|$. Thus, this proves once again that the two coincides when $\displaystyle 1=r=|z|$. As for your question $\displaystyle \mathbb{S}^1$ is just the unit circle $\displaystyle |z|=1$.
• Dec 26th 2010, 03:52 PM
aonin
there was a question regarding graphing 1/z and conj z and they were the same line
then i realised that through demoivre's theorem that the conj and 1/z could be proved to be equal

however when i realised the denominator in 1/z where eg. z=2+3i the result is (2+3i)/13 which is consistent to the rule z^-1=conj z/(mod z)^2

so why are there 2 answers by using 2 methods (demoivre's theorem and realising the denominator), or am i misunderstanding something?
• Dec 26th 2010, 04:00 PM
aonin
yep i get what you mean now. i just forgot about the modulus when doing (rcistheta)^-1=r^-1 X cis-theta. thanks for clearing that up
• Dec 26th 2010, 04:04 PM
dwsmith
You can't graph Complex Numbers like Real Numbers since you are dealing with 4 dimensions.

$\displaystyle w=f(\bar{z}) \ \ \{w\in\mathbb{C}:w\neq 0\}$

$\displaystyle \displaystyle\mbox{Circle:} \ w=f\left(\frac{1}{z}\right) \ \ \left\{w\in\mathbb{C}:w\neq 0 \ |w|=\left|\frac{1}{r}\right|\right\}$

$\displaystyle \displaystyle x=k: \ \ w=f\left(\frac{1}{z}\right) \ \ \left\{w\in\mathbb{C}:w\neq 0 \ \left|w-\frac{1}{2k}\right|=\left|\frac{1}{2k}\right|\righ t\}$

$\displaystyle \displaystyle y=k: \ \ w=f\left(\frac{1}{z}\right) \ \ \left\{w\in\mathbb{C}:w\neq 0 \ \left|w+\frac{1}{2k}\mathbf{i}\right|=\left|\frac{ 1}{2k}\right|\right\}$

Extended Complex Plane includes w=0.
• Dec 26th 2010, 06:10 PM
mr fantastic
Quote:

Originally Posted by aonin
there was a question regarding graphing 1/z and conj z and they were the same line
then i realised that through demoivre's theorem that the conj and 1/z could be proved to be equal

however when i realised the denominator in 1/z where eg. z=2+3i the result is (2+3i)/13 which is consistent to the rule z^-1=conj z/(mod z)^2

so why are there 2 answers by using 2 methods (demoivre's theorem and realising the denominator), or am i misunderstanding something?

Again we have a situation where, if the original question had been posted rather than some misleading/incorrect interpretation of it, a lot of time and effort would have been saved ....