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?

2. 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 $z^{-1}=\bar{z}$ since as you noted for $z\ne 0$ you have that $z^{-1}=\frac{\bar{z}}{|z|^2}$ from where it quickly follows that $\bar{z}=z^{-1}\implies z\in\mathbb{S}^1$ and the converse can be checked quite readily.

3. 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

4. What were you asked to do originally?

You have all these work based on something wrong.

Are you asked to prove or disprove if $z^{-1}=\bar{z} \ \mbox{?}$ If so, an example of why it doesn't suffices which you have. If not, what are you trying to do?

5. There are several mistakes in your OP.
Consider this fact: $z^{ - 1} = \dfrac{1}{z} = \dfrac{{\overline z }}{{\left| z \right|^2 }}$.

What does that do to what you posted?

6. 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
I don't see how DeMoivre's theorem: $\displaystyle (\cos(\theta)+i\sin(\theta))^n=\cos(n\theta)+i\sin (n\theta)$, implies this. Aha! I bet I know where you got messed up! You can get from DeMoivre's theorem that $\left(\cos(\theta)+i\sin(\theta)\right)^{-1}=\cos\left(-1\theta\right)+i\sin(-\theta)=\cos(\theta)-i\sin(\theta)=\overline{\cos(\theta)+i\sin(\theta) }$ and you had a fuzzy idea that every number can be represented as something 'like' $\cos(\theta)+i\sin(\theta)$ but you forgot that it's actually $r(\cos(\theta)+i\sin(\theta))$ where $r=|z|$. Thus, this proves once again that the two coincides when $1=r=|z|$. As for your question $\mathbb{S}^1$ is just the unit circle $|z|=1$.

7. 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?

8. 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

9. You can't graph Complex Numbers like Real Numbers since you are dealing with 4 dimensions.

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

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

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

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

Extended Complex Plane includes w=0.

10. 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 ....