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: