find the general value of log(4+3i)
Answers
Answer:
4 + 3i is 4 – 3i.
Step-by-step explanation:
Next we need the product of these two complex numbers. (If not then you can multiply the two binomials the hard way)
( X + Y ) x (X - Y) = X^2 - Y^2.
Letting Y be the imaginary component, ( 4 + 3i ) x ( 4 - 3i ) = 4^2 - [ 3i]^2
= 16 - 9 ( i^ 2 ) = 16 + 9 = 25
Recalling that i ^ 2 = -1 by definition, which changes the sign of the second component.
Since our result is a real rather than a complex number, standard form is a bit excessive, but you could express it as --
25 + 0i, that is zero times i , implying that there is no imaginary term.
Given:
log(4+3i)
To find:
The general value of log(4+3i)
Solution:
Here, the argument(ø) of our complex number of (4+3i) is,
tan-1 (3/4) ( Since it is tan inverse).
From trigonometry,
we have,
sinø = 3/5
cosø = 4/5.
The modulus of (4+3i) is 5.
Hence,
(4+3i) = 5(4/5 + 3/5i).
(4+3i) = 5(cosø + isinø).
Using Euler's Form,
cosø + isinø = e.(i.ø).
Therefore,
log(4+3i) = log[5.e..(i.ø)]
By logarithm property,
log[5.e.(i.ø)] = log5 + log[e..(i.ø)]
log5 + i.ø
Substituting the value of ø
log5 + i.tan-1(3/4)
Therefore, the real part of the log(4+3i) is log5.