Question

z=2(cosπ/3+isinπ/3)
Calculate ( z̅ )^-1

Answer 1

do you know how to find conjugate of a complex number?

Answer 2

Yes but i want working.

Answer 3

z̅ =2(cosπ/3-isinπ/3)
( z̅ )^-1=1/2(cosπ/3-isinπ/3)
Is this solution right?

Answer 4

complex conjugates happen in paris,

\[z=a \pm bi\]

we denote \[z̅\] as its complex conjugate of z

Answer 5

pairs

Answer 6

here, z is the positive pair

therefore z̅ will be the negative value

Answer 7

i must have dinner. someone else can elaborate.

Answer 8

@Jhannybean @zepdrix

Answer 9

@Directrix

Answer 10

Hey Nurali :)

z̅ =2(cosπ/3-isinπ/3)
( z̅ )^-1=1/(2(cosπ/3-isinπ/3))

What you wrote above ^ looks correct,
but we would like to go a little further and bring the cosines and sines to the numerator.

Answer 11

\[\large\rm z=2\left(\cos\frac{\pi}{3}+i \sin\frac{\pi}{3}\right)\]\[\large\rm \overline{z}=2\left(\cos\frac{\pi}{3}-i \sin\frac{\pi}{3}\right)\]\[\large\rm \bar{z}^{-1}=\frac{1}{2\left(\cos\frac{\pi}{3}-i \sin\frac{\pi}{3}\right)}=\frac{1}{2}\left(\cos\frac{\pi}{3}-i \sin\frac{\pi}{3}\right)^{-1}\]

Answer 12

We could apply De'Moivre's Theorem from this step.
Do you remember that one? :)

Answer 13

\[\large\rm \left[\cos(\theta)+i \sin(\theta)\right]^{n}=\cos(n \theta)+i \sin(n \theta)\]Do you see how we could apply that to our problem here?

Answer 14

can u explain?

Answer 15

I used exponent rule to bring the cosine and sine up to the numerator.
That placed a -1 power on the whole thing.
I'll apply De'Moivre's Theorem to bring the -1 power into the angle,\[\large\rm \bar{z}^{-1}=\frac{1}{2}\left[\cos\frac{\pi}{3}-i \sin\frac{\pi}{3}\right]^{-1}\]\[\large\rm \bar{z}^{-1}=\frac{1}{2}\left[\cos\left(-1\cdot\frac{\pi}{3}\right)-i \sin\left(-1\cdot\frac{\pi}{3}\right)\right]\]\[\large\rm \bar{z}^{-1}=\frac{1}{2}\left[\cos\left(-\frac{\pi}{3}\right)-i \sin\left(-\frac{\pi}{3}\right)\right]\]

Answer 16

From here, we'll apply properties of sine and cosine.

Sine is an `odd function`, meaning \(\large\rm sin(-x)=-sin(x)\)
While Cosine is an `even function`, so \(\large\rm cos(-x)=cos(x)\)

Answer 17

So our \(\large\rm \cos\left(-\dfrac{\pi}{3}\right)\) will turn into \(\large\rm \cos\left(\dfrac{\pi}{3}\right)\)

while our \(\large\rm \sin\left(-\dfrac{\pi}{3}\right)\) will turn into \(\large\rm -\sin\left(\dfrac{\pi}{3}\right)\)

Answer 18

\[\large\rm \bar{z}^{-1}=\frac{1}{2}\left[\color{orangered}{\cos\left(-\frac{\pi}{3}\right)}-i \color{royalblue}{\sin\left(-\frac{\pi}{3}\right)}\right]\]Becomes,\[\large\rm \bar{z}^{-1}=\frac{1}{2}\left[\color{orangered}{\cos\frac{\pi}{3}}-i \color{royalblue}{\left(-\sin\frac{\pi}{3}\right)}\right]\]Further simplifying to,\[\large\rm \bar{z}^{-1}=\frac{1}{2}\left[\cos\frac{\pi}{3}+i \sin\frac{\pi}{3}\right]\]

Answer 19

Hmm this looks a whole lot like our z that we started with, doesn't it?
Let's do something clever, let's multiply by 2,
we'll divide by 2 as well to offset that.\[\large\rm \bar{z}^{-1}=\frac{1}{2}\cdot\frac{1}{2}\cdot\color{red}{2\left[\cos\frac{\pi}{3}+i \sin\frac{\pi}{3}\right]}\]
So check this out!! Kinda neat!\[\large\rm \bar{z}^{-1}=\frac{1}{2}\cdot\frac{1}{2}\cdot\color{red}{z}\]

Answer 20

What do you think?
Too confusing?

I'll give you a chance to digest all of that lol.

Answer 21

No I understand thanks