find the general and principal value of (i)i
Answers
Using the laws of logarithms,
log
(
−
1
+
i
)
−
(
−
1
−
i
)
=
log
(
−
1
+
i
−
1
−
i
)
Amplify the complex number by its conjugate:
−
1
+
i
−
1
−
i
⋅
−
1
+
i
−
1
+
i
=
(
−
1
+
i
)
2
1
2
−
i
2
=
1
−
2
i
−
1
1
+
1
=
−
2
i
2
=
−
i
∴
log
(
−
1
+
i
−
1
−
i
)
=
log
(
−
i
)
The base of the logarithm hasn't been stated as of yet, but I assume it is the natural logarithm. From now onwards, I will use
ln
to denote it.
We want to find:
ln
(
−
i
)
We know that the Polar form of a complex number is
z
=
r
(
cos
θ
+
i
sin
θ
)
=
r
e
i
θ
⇒
ln
z
=
ln
r
+
i
θ
Let
z
=
−
i
.
ln
(
−
i
)
=
ln
r
+
i
θ
=
ln
|
−
i
|
+
i
θ
=
i
θ
Returning back to our trigonometric form:
−
i
=
cos
θ
+
i
sin
θ
0
−
1
i
=
cos
θ
+
i
sin
θ
⇒
{
cos
θ
=
0
sin
θ
=
−
1
The principal value of
θ
is the smallest positive one, hence:
θ
=
3
π
2
is the principal value.
As such, the general value of
θ
is:
θ
=
3
π
2
+
2
k
π
,
k
∈
Z
.
Therefore:
ln
(
−
i
)
=
ln
(
−
1
+
i
)
−
ln
(
−
1
−
i
)
=
i
θ
=
i
(
3
π
2
+
2
k
π
)
Principal value:
ln
(
−
1
+
i
)
−
ln
(
−
1
−
i
)
=
i
3
π
2
General value:
ln
(
−
1
+
i
)
−
ln
(
−
1
−
i
)
=
i
(
3
π
2
+
2
k
π
)