Write lnz=ln∣z∣+iargz and ez=ex(cosy+isiny) with z=x+iy.
a) lnz is purely imaginary ⇔Re(lnz)=ln∣z∣=0⇔∣z∣=1: the unit circle. Moreover Im(lnz)=argz=6π is the ray from the origin with inclination 6π.
b) ez real positive ⇔y=2kπ; real negative ⇔y=π+2kπ; imaginary ⇔y=2π+kπ; real part = imaginary part ⇔cosy=siny⇔y=4π+kπ.
c) On ∣z∣=1 we have lnz=ln1+iargz=iargz with argz∈(−π,π]: the circle becomes the segment of the imaginary axis from −iπ to iπ.
a) ∣z∣=1 and ray argz=6π;c) segment [−iπ,iπ]