express log[log(costhita+ isinthita)] in form of A +iB
Answers
Answer:A
=
ln
π
B
=
π
/
2
+
2
n
π
where
n
is an integer.
Explanation:
Instead of a vague
log
, let's use the natural logarithm so things simplify nicely.
We have a complex number
z
=
ln
ln
(
cos
π
+
i
sin
π
)
for which we want to find the real part,
A
and the imaginary part,
B
respectively.
What we have inside of the composed
ln
function resembles Euler's identity, which states that
e
i
x
=
cos
x
+
i
sin
x
In our particular case,
x
=
π
, hence
cos
π
+
i
sin
π
=
e
i
π
If we plug this in our original expression, we get a simplified form:
ln
ln
(
cos
π
+
i
sin
π
)
=
ln
ln
e
i
π
=
ln
i
π
This can be expanded using the properties of logarithms:
ln
i
π
=
ln
i
+
ln
π
All that is left is finding
ln
i
.
We already know a relation between the base of the natural logarithm,
e
, and complex numbers. To find out
ln
i
, suppose there exists
α
such
i
=
e
i
α
.
If we write these in their complex number form we can see the similarity:
i
=
0
+
1
⋅
i
e
i
α
=
cos
α
+
i
sin
α
⇒
{
cos
α
=
0
sin
α
=
1
One solution to this would be
α
=
π
/
2
, and since the trigonometric functions are periodic with period
ρ
=
2
n
π
for any integer
n
, the set of solutions is
α
=
π
2
+
2
n
π
Our complex number can be yet again written as
z
=
ln
i
+
ln
π
=
ln
e
i
α
+
ln
π
=
ln
π
+
α
i
∴
z
=
ln
π
+
i
(
π
2
+
2
n
π
)
,
n
∈
Z
We have solved the question in discussion:
{
A
=
ln
π
B
=
π
/
2
+
2
n
π
Answer linkA
=
ln
π
B
=
π
/
2
+
2
n
π
where
n
is an integer.
Explanation:
Instead of a vague
log
, let's use the natural logarithm so things simplify nicely.
We have a complex number
z
=
ln
ln
(
cos
π
+
i
sin
π
)
for which we want to find the real part,
A
and the imaginary part,
B
respectively.
What we have inside of the composed
ln
function resembles Euler's identity, which states that
e
i
x
=
cos
x
+
i
sin
x
In our particular case,
x
=
π
, hence
cos
π
+
i
sin
π
=
e
i
π
If we plug this in our original expression, we get a simplified form:
ln
ln
(
cos
π
+
i
sin
π
)
=
ln
ln
e
i
π
=
ln
i
π
This can be expanded using the properties of logarithms:
ln
i
π
=
ln
i
+
ln
π
All that is left is finding
ln
i
.
We already know a relation between the base of the natural logarithm,
e
, and complex numbers. To find out
ln
i
, suppose there exists
α
such
i
=
e
i
α
.
If we write these in their complex number form we can see the similarity:
i
=
0
+
1
⋅
i
e
i
α
=
cos
α
+
i
sin
α
⇒
{
cos
α
=
0
sin
α
=
1
One solution to this would be
α
=
π
/
2
, and since the trigonometric functions are periodic with period
ρ
=
2
n
π
for any integer
n
, the set of solutions is
α
=
π
2
+
2
n
π
Our complex number can be yet again written as
z
=
ln
i
+
ln
π
=
ln
e
i
α
+
ln
π
=
ln
π
+
α
i
∴
z
=
ln
π
+
i
(
π
2
+
2
n
π
)
,
n
∈
Z
We have solved the question in discussion:
{
A
=
ln
π
B
=
π
/
2
+
2
n
π
Answer link
Step-by-step explanation: