A Complex Logarithmic Equation | Problem 410

[@scottleung9587]

I also got -e^pi.

[@crazyAngol]

Wolfram Alpha also got -e^pi

[@brutusmag1]

There's a much simpler way: z=exp(%pi+%pi*%i)=exp(%pi)*exp(%pi*%i)=exp(%pi)*(-1)=-exp(%pi) and done

[@bjornfeuerbacher5514]

That's the second method in the video, starting at 8:05.

[@brutusmag1]

@@bjornfeuerbacher5514 You're right, sorry for not watching the video completely.

[@trojanleo123]

Just my two cents on this - It's good that you show multiple methods to solve a problem and that's important. I think you should also reinforce forcefully in your videos which is the 'advisable' method to solve the problem. It reinforces the notion that just because you can do something a certain way doesn't mean that you should. For e.g. while one can solve this problem using method 1, one should never use that method to solve such a problem. Just my 2 cents.

[@aplusbi]

Good thinking!

[@CriticSimon]

I think it’s better if the viewer decides what’s easier or more convenient for them, not the instructor.

[@trojanleo123]

@CriticSimon Doesn't have to be either/or. It can be both the instructor and the viewer. But That's alright we are allowed to disagree.

[@CriticSimon]

@ Thanks. I agree with you now ☺️

[@seanfraser3125]

-e^pi

[@Mediterranean81]

-e^π Also is that a prerecorded voice ?

[@aplusbi]

No

[@Mediterranean81]

@@aplusbicongrats on getting your voice back

[@angelommv]

lnz = pi + i*pi 1) => e^(lnz) = e^(pi + i*pi) => z = [e^pi] * [e^i*pi] => z = (e^pi)*(-1) => z = -e^pi 2) ln(z)= pi + i*pi but ln(x*y) = lnx + lny if x*y=z => lnx = pi ; lny = i*pi easy, x=e^pi and y=e^(i*pi)=-1 x*y = z = -e^pi

[@brutusmag1]

Yes, the 1) is the most simple, and the second one is also clever.

[@DavidChristman-q7l]

lnz=pi+ipi e^lnz=e^(pi+ipi) z=e^(pi+ipi) z=(e^pi)(e^ipi) z=(e^pi)(-1) z=-e^pi

[@NadiehFan]

Your error with your first method is that you took arctan(b/a) to be π, this is incorrect. The arctangent of a real number is always on the open interval (−½π, ½π) whereas the principal argument of a nonzero complex number can have any value on the interval (−π, π]. This implies that arctan(b/a) only gives the principal argument for a complex number a + bi in the _right half_ of the complex plane.

[@trojanleo123]

z = -e^π

[@ВикторПоплевко-е2т]

Method 1 is MUCH easier though

[@Peter-rm7io]

Well if you have used the Euler formula in the beginning, it would have taken like five seconds.

[@CriticSimon]

That’s not the point! Most people need to see the steps involved. Not everyone is as genius as you are 😊

[@mathmachine4266]

z=-e^π