Calculus 2: Complex Numbers & Functions (21 of 28) Finding the nth Root: Example

In this video I will find z^1/3=? given z=27[(pi)+isin9pi)].

Next video in the series can be seen at:

[lordyabo]

Sir does this technique should only be use if the exponent of z is a fraction? and if the exponent of z is a whole number the technique in video 19 is the one that should be use? Thanks!

[theAplaBetman]

Thank you so much, Professor!! This helps a lot

[countzero]

I think an interesting thing to add to this is that when k = 2 the angle (5*pi)/3 is in Q4 which means it's the same as the angle (- pi/3) which is a variant of the result when k = 0:

k = 2 => 3[cos (-pi/3) + i sin (-pi/3)] = 3[cos (pi/3) - i sin (pi/3) ]

k = 0 => 1.5 + 2.6i
k = 2 => 1.5 - 2.6i

[EP-rqpn]

Thank you for this Mr van B. I'm having trouble wrapping my mind around needing to list all applicable n*2pi solutions. I understand where that came from = because of the periodicity, but by this logic, wouldn't we also need to list the additional solutions when we raise z to an integer exponent? Wouldn't that still apply for all n values up to the exponent? Why does it matter just for the roots? Is there a more in-depth look into the intuition for this concept? I can't see it

[matricmasterclass]

Hi Michel


I was wondering, when do you know when to stop? If you were to continue substituting k, would you just get the same values over and over again?

[shivanshutiwari]

sir your videos help me to preparation of IIT jee thanks sir

[anasghaffar]

Thank You! Helped alot, Also if you add another video clarifying what the roots are actually and draw a clear picture, will be helpful!