## Ahsome one year ago How to find conjugate of complex number

1. ahsome

Given $$w=2\text{cis}\left(-\dfrac{\pi}{5}\right)$$ Find $$\overline{w}$$ Is there anyway of solving this without turning into $$x+yi$$, then finding conjugate, then back to $$\text{cis}$$ form?

2. ahsome

@ganeshie8?

3. amistre64

what does W multiplied by it conjugate result in?

4. ahsome

$$x^2+y^2$$?

5. ahsome

Also, imaginary part = 0

6. amistre64

seems fair to me what is the 2 part in W ?

7. ahsome

The magnitude

8. amistre64

for 2 complex numbers (if memory serves) (n,a)(m,b) = (nm,a+b) n,m being magnitudes, and a,b being angles

9. ahsome

Yup :)

10. amistre64

cos(0) + sin(0) = 1 so a+b = 0 right? or am i going down a rabbit hole?

11. anonymous

I think you can get the conjugate by using the negative theta rcis(-theta)

12. ahsome

Yes, I understand that

13. ahsome

And @Harindu, reallly?

14. ahsome

Well, i guess that would make sense

15. ahsome

Or would it be thetha - $$\pi$$?

16. ahsome

nvm

17. anonymous

sorry I haven't touched complex numbers for sometime. But a + bi = r cis(theta) Where: a = rCos(theta) b = rSin(theta ) cos only b gets negative if theta is negative it works out

18. ahsome

Yup, I get that. Thanks @Harindu :)

19. anonymous

You are welcome

20. ahsome

How would you solve $\frac{1}{w}$ tho?

21. ahsome

@Harindu?

22. amistre64

(n,a)^k = (n^k, ka)

23. ahsome

In this case, would it be -1?

24. amistre64

after all (n,a)(n,a) = (nn,a+a) yes, in this case k=-1

25. anonymous

hmm turn into a complex number first . I mean get the denominator to be real so I think you have to use x + yi form . I dnt remember sorry

26. ahsome

Wait, isn't 1 = $$1\times cis(\theta)$$?

27. amistre64

w1/w2 (n,a) (m^(-1), -b) = (n/m, a-b)

28. ahsome

Then, you can minus them. $$\text{Let }1=z = 1\times cis(\theta)$$ $$\text{Let }w = 4\times cis\left(\dfrac{-\pi}{5}\right)$$ $\frac{1}{w}$$\dfrac{4}{1}\times cis\left(\dfrac{-\pi}{5}-1\right)$Does that work?

29. ahsome

Althought it would be 1/4 and $$1-\frac{-\pi}{5}$$

30. amistre64

$\frac1{4cis(-\frac{\pi}{5})}$ $\frac{1}{4}cis(\frac{\pi}{5})$

31. ahsome

:D

32. ahsome

THANK YOU :)

33. amistre64

$(4,-\frac{\pi}{5})^{-1}\implies(4^{-1},-(-\frac{\pi}{5}))$

34. amistre64

youre welcome

35. ahsome

No problem. For future reference, if it ever occured that: $\frac{x}{w}$ We would simply say that $\frac{x}{w}=\frac{x\times cis(0)}{w}$ Then do $\frac{x}{|w|}\times cis\left(0-\arg(w)\right)$?

36. ahsome

If x is a real number and w is a complex number?

37. amistre64

yep, seems fair to me

38. ahsome

Thank you :)