Thanks for the video! Clear and concise. Also, for those that are curious about the formula, it is pretty simple to derive when you represent the complex number in exponential form: Let A = magnitude of a complex number, and theta = phase of the complex number: sqrt( A * exp(j * theta) ) => (A * exp(j * theta))^(1/2) Using the laws of exponents to simplify: => A^(1/2) * exp(j * theta / 2) => sqrt(A) * exp(j * theta / 2) Converting back to phasor form yields: sqrt(A) ∠ (theta/2) I like exponential notation in this case because it is easier to work with than rectangular notation, which can save you time on exams.
Thank you very much. Thanks to you, I was able to solve this problem that I have been dealing with for a long time. I hope you will always be very happy.
I am glad you liked it...yes you can divide by 3 for cube root and 4 for root 4 but that will give only 1 solution out of 3 in cube root and 4 in 4th root...for others you will have to add 2n pi to the original angles and divide by 3 or 4 to get different angles
You can calculate the same things in 9991cw as well...just search for 991cw complex numbers and you will understand the standard functions and then apply the same method
You can divide by 3 for cube root and 4 for root 4 but that will give only 1 solution out of 3 in cube root and 4 in 4th root...for others you will have to add 2n pi to the original angles and divide by 3 or 4 to get different angles
Thanks for the video! Clear and concise. Also, for those that are curious about the formula, it is pretty simple to derive when you represent the complex number in exponential form:
Let A = magnitude of a complex number, and theta = phase of the complex number:
sqrt( A * exp(j * theta) )
=> (A * exp(j * theta))^(1/2)
Using the laws of exponents to simplify:
=> A^(1/2) * exp(j * theta / 2)
=> sqrt(A) * exp(j * theta / 2)
Converting back to phasor form yields:
sqrt(A) ∠ (theta/2)
I like exponential notation in this case because it is easier to work with than rectangular notation, which can save you time on exams.
Nice explanation...i will pin this comment so others will find it on the top
Thanks m8
Thank you very much. Thanks to you, I was able to solve this problem that I have been dealing with for a long time. I hope you will always be very happy.
Happy to help
Thank you so much sir.This technique will greatly help me in my upcoming exam❤❤
Happy to help
Superb. You have made it look easy.
Thanks a lot 😊
درود بر شما!
Thank you
Thank you, that was really helpful 🙏👍🏼
I appreciate that
For God sake, i've watched this Video more than 100 times,, Everytime i understand it. But after a few week i forget it Again.. 😢
Thats unfortunate...but i am happy you are watching it again and again :D
Hello thank for this, I have a question, how about if its for cuberoot? Do I divide it by 3? And divide by 4 if rooted to 4?
I am glad you liked it...yes you can divide by 3 for cube root and 4 for root 4 but that will give only 1 solution out of 3 in cube root and 4 in 4th root...for others you will have to add 2n pi to the original angles and divide by 3 or 4 to get different angles
I see, thank you
Thanks for the video sir.
Welcome
may i askk? why the angle is being divided by 2 ?
because we are taking square root...if it was cue root we would divide by 3
Can we use equation editor in cm1 exam from IAI ? Or not ?
Yes we can
Hello what if the complex number is raised by 3/2?
You can cube it first and then find the root
Please please please make a video how to find square root of a complex number using FX 991cw calculator
You can calculate the same things in 9991cw as well...just search for 991cw complex numbers and you will understand the standard functions and then apply the same method
Very simple using demuaver method
Thanks for the comment but i couldn't find demuaver method...if you could share more information, it would be helpful
@@tooirrational i meant you used demuaver, I.didn't find de muaver in the calculator as well . sorry for miss understanding.
No problem 😊
But it's only one root, what about the other root?
You can divide by 3 for cube root and 4 for root 4 but that will give only 1 solution out of 3 in cube root and 4 in 4th root...for others you will have to add 2n pi to the original angles and divide by 3 or 4 to get different angles
Thank you
You're welcome
wow. thank you
Happy to help
Thank you sir
Happy to help
Thanks a lot 😊
Happy to help
Thanks❤
You're welcome 😊
Sharma Ji sukriya
happy to help
Legend!!!
Glad it was helpful!
Tq sir😊
Welcome 😊
Thank you so much! TT_TT
Happy to help
Take it❣️
Thank You
Why is it showing me an error
You will have to be more specific