Question

If log (x -iy)=A+iB, where A and B are real then
0​

Answer 1

Answer:

Dear,what is your question?

please, provide full question.

Step-by-step explanation:

Because, I think in that question, something is missing.

Answer 2

Step-by-step explanation:

Log (a +i*b)=ω=x+ i*y. say

∴ By laws of logarithm

e

$e^{x+iy}$  =a +i* b

∴$e^{x}$ * $e^{iy}$  =a +i*b

∴$e^{x}$ * *[cis .y]=a +i*b

⇒($e^{x}$ *cos(y))+($e^{x}$* sin(y))*i=a +ib

equating the real & imaginary part,

a=$e^{x}$ *cos(y)------(1)

b=$e^{x}$ *sin(y)------(2)

To find  ′x ′ , square (1) & (2) & add them

∴$e^{2x}$* ($sin ^{2}$ y + $cos^{2}$ y) = $a^{2}$ + $b^{2}$

∴$e^{2x}$ = $a^{2}$ + $b^{2}$