An Interesting Exponential Equation | Problem 230

You can raise both side by i

Very cool!

Representing z as ae^(i*theta), we get (ae^(i*theta))^(-iz)=i, which simplifies to ae^(-ii*theta*z)=i or more simply ae^(theta*z)=i. As i has a magnitude of 1, we know a=1. Representing i in polar form as e^(i*pi/2), we get e^(theta*z)=e^(i*pi/2). Setting the exponents equal to one another, theta*z=(pi/2)i. Presuming a non-imaginary argument, rearranging we get theta/(pi/2)=z/i, which tells us that z rotated in the complex plane pi/2 clockwise is a real number, such that z has a real part of 0 with magnitude 1. The only options for z are thus i and -i, with theta being pi/2 or -pi/2 respectively. i^(-ii)=i and (-i)^(-i*-i)=(-i)^-1=i.

Super

ie^W(0) if I did not wrong any calculations. Gonna watch the video Edit: wow, it turned out cooler than I expected 8).

Hey there! Had an idea for a problem. The other day you did a system of linear equations with complex variables and complex coefficients. And you did it a few different ways. And you asked if there were other ways. How about doing it as an augmented matrix? Instead of the system: 2z + iw = 4 - i iz - 3w = 2 + 2i You could make an augmented matrix and then perform the usual linear algebra shenanigans to solve it that way: | 2 i : 4-i| |i -3 : 2+2i| (Can't make a nice, neat matrix in YouTube comments, but you get the idea.

W(-1)

Very interesting and easy eq

And the other solutions? 😉

ln(z^-iz) = lni -izlnz = lni zlnz = -lni / i zlnz = ilni therefore z=i