Find all the values of log(3-2i)
Well, log(3-2i) is complex logarithm, so let's just say I'm log-o-scopic and I'll give you the values. Remember, this is going to be a wild logarithmic ride!
The complex number 3-2i can be represented as r * e^(iθ), where r is the magnitude of the complex number and θ is the angle it makes with the positive x-axis when plotted in the complex plane.
To find the values of log(3-2i), we need to find the values of r and θ. The magnitude of the complex number can be calculated using the formula sqrt(a^2 + b^2), where a and b are the real and imaginary parts, respectively.
For 3-2i, the magnitude is sqrt((3^2) + (-2^2)) = sqrt(9 + 4) = sqrt(13).
Now, let's find the value of θ. We can use the arctan function to find the angle formed between the complex number and the positive x-axis. The formula is θ = arctan(b/a), where a is the real part and b is the imaginary part.
For 3-2i, θ = arctan((-2)/(3)) = -0.588 radians (approximately).
So, we have r = sqrt(13) and θ = -0.588 radians.
Therefore, the values of log(3-2i) can be written as log(sqrt(13)) + i * (-0.588) + 2 * π * i * k, where k is an integer.
You can see that there are infinitely many values for log(3-2i), each differing by a multiple of 2πi. So, you're in for a logarithmic circus with an endless parade of complex logarithmic values! Enjoy the show!
To find the values of log(3-2i), we can use the polar form of a complex number.
Step 1: Find the modulus (magnitude) of the complex number.
The magnitude of 3-2i can be found using the formula: |z| = sqrt((a^2) + (b^2)), where a is the Real part (3) and b is the Imaginary part (-2).
Thus, |3-2i| = sqrt((3^2) + (-2^2)) = sqrt(9 + 4) = sqrt(13).
Step 2: Find the argument of the complex number.
The argument (θ) of 3-2i can be found using the formula: θ = arctan(b/a), where a is the Real part (3) and b is the Imaginary part (-2).
Thus, θ = arctan(-2/3) ≈ -33.69 degrees.
Step 3: Write the complex number in polar form.
By using the magnitude and argument, we can express 3-2i in polar form as:
3-2i = sqrt(13) * [cos(-33.69 degrees) + i sin(-33.69 degrees)].
Step 4: Express log(3-2i) in exponential form.
Using Euler's formula, e^(ix) = cos(x) + i sin(x), we can rewrite the polar form as:
3-2i = sqrt(13) * e^(-33.69 degrees * i).
Step 5: Apply the logarithmic property.
To find log(3-2i), we can apply the logarithmic property: log(a * b) = log(a) + log(b).
So, log(3-2i) = log(sqrt(13)) + log(e^(-33.69 degrees * i)).
Step 6: Evaluate each part separately.
log(sqrt(13)) is the logarithm of the modulus (magnitude) of the complex number, which is approximately 0.832.
log(e^(-33.69 degrees * i)) is the logarithm of the exponential form, which equals -33.69 degrees * i.
Therefore, the values of log(3-2i) are approximately:
0.832 - 33.69 degrees * i.
To find the values of log(3 - 2i), we can use the logarithmic properties and the polar form of a complex number.
Step 1: Write the complex number in polar form.
The polar form of a complex number is given by r(cos θ + i sin θ), where r is the magnitude of the complex number and θ is the argument or angle.
In this case, the complex number is 3 - 2i. To find its polar form, we need to calculate the magnitude (r) and the argument (θ).
The magnitude (r) can be found using the formula:
r = √(a^2 + b^2), where a = 3 and b = -2
r = √(3^2 + (-2)^2)
r = √(9 + 4)
r = √13
To find the argument (θ), we can use the inverse tangent function:
θ = tan^(-1)(b/a), where a = 3 and b = -2
θ = tan^(-1)(-2/3)
Step 2: Write the complex number in exponential form.
The exponential form of a complex number in polar form is given by re^(iθ), where r is the magnitude and θ is the argument.
In this case, the exponential form is √13 * e^(i * tan^(-1)(-2/3)).
Step 3: Take the natural logarithm of the exponential form.
log(√13 * e^(i * tan^(-1)(-2/3))) = log(√13) + log(e^(i * tan^(-1)(-2/3)))
Using the logarithmic property log(ab) = log(a) + log(b), the equation simplifies to:
log(√13) + log(e^(i * tan^(-1)(-2/3)))
Step 4: Evaluate the logarithmic expressions.
The first term, log(√13), can be simplified by using the property log(x) = y if and only if e^y = x:
e^(log(√13)) = √13
The second term, log(e^(i * tan^(-1)(-2/3))), can be simplified using the fact that log(e) = 1:
log(e^(i * tan^(-1)(-2/3))) = i * tan^(-1)(-2/3)
Therefore, the values of log(3 - 2i) are √13 + i * tan^(-1)(-2/3).
I will assume you want:
ln(3-2i)
I will assume that you are familiar with De Moivre's Theorem
and the definition for r e^(θi)
let z = ln(3-2i)
then e^z = 3-2i
= √13 e^(5.695i) <---- from r^2 = 3^2 + (-2)^2 and tanθ = -2/3 in quadrant IV
= (e^(ln √13) * e^(5.695i) <----- log property: e^(ln k) = k
= e^( ln √13 + 5.695i)
so if e^z = e^( ln √13 + 5.695i)
z = ln √13 + 5.695i