The real part of (1-i)-i is
e-π4cos12log2
-e-π4sin12log2
eπ4cos12log2
e-π4sin12log2
Compute the real part:
Let z=(1-i)-i
Taking log on both sides,
logz=-ilog1-i logax=xloga
⇒logz=-ilog2cosπ4-isinπ4
⇒logz=-ilog2.e-iπ4
⇒logz=-i12log2+loge-iπ4 loga.b=loga+logb
⇒logz=-i12log2-iπ4
⇒logz=-i2log2-π4
⇒ z=e-π4e-i2log2
Taking real part ,
⇒ z=e-π4cos(12log2)
Hence, option (A) is the correct option.