Question

What is the principal value of log(i^1/4)

Answer 1

Step-by-step explanation:

Letz=1−i

|z|=1+1−−−−√=2–√arg z=−tan−1(1)=−π4

∴z=2–√cis(−π4)=2–√e−π4

z4i=(2–√)4ie−π4i×(4i)=22ieπ(1)

Lety=22i⟹lny=ln22i=2iln(2)

⟹y=ei(2ln2)=cos(2ln2)+isin(2ln2)

putting value of y in (1)

z4i=eπ(cos(2ln2)+isin(2ln2))

⟹(1−i)4=eπ(cos(2ln2)+isin(2ln2))