Can We Solve A Very Exponential Equation? | Problem 243

[@ShauyanOfficial13]

no wolfram alpha this video??

[@ismailfahmy8041]

"makes sense?" No man I forgot half of my A-Level maths 😭

[@aplusbi]

No worries you can always get back to it

[@Michael_L_]

At 3:50 could we not have multiplied both sides by -i and then solved for i^(z-1) ?

[@mcwulf25]

Presumably. Does it give the same answer?

[@DavyCDiamondback]

I already know from inspection that i^z = [2k(pi) + (pi)/2]i which I'm sure I can solve with i^z=e^(zlni). We know the modulus of [2k(pi) + (pi)/2]i: = pi/2 for k >= 0, and = -pi/2 for k < 0, and the magnitude: = 2k(pi) + (pi)/2 for all k...

[@DavyCDiamondback]

I'll just watch the video, this is annoying without paper