Find all complex numbers z satisfying the equation e 6
iz = 3i. Give your answer in
Cartesian form, and indicate them on the complex plane.
z = 1/12 (4πn - log(9) i)
To solve the given equation e^(6iz) = 3i, we can start by expressing the complex numbers in their Cartesian form.
First, let's convert the complex number 3i to Cartesian form. In the Cartesian form, a complex number is represented as z = x + yi, where x and y are real numbers. For 3i, the real part (x) is 0, and the imaginary part (y) is 3. Therefore, 3i can be written as 0 + 3i.
Next, let's write the complex number iz in Cartesian form. Since z = x + yi, iz = (0 + 1i) * (x + yi) = -y + xi.
Now, we have the equation e^(6(-y+xi)) = 3i, which simplifies to e^(6xi - 6yi) = 3i.
To continue, we need to express e^(6xi - 6yi) in polar form. In the polar form, a complex number is represented as z = r * e^(iθ), where r is the magnitude (distance from the origin) and θ is the argument (angle with the positive x-axis).
Using Euler's formula, we can write e^(6xi - 6yi) = e^(-6y) * e^(6xi) = e^(-6y) * (cos(6x) + i*sin(6x)).
Comparing this expression with e^(6xi - 6yi) = 3i, we can deduce that e^(-6y) = 3 and cos(6x) + i*sin(6x) = i.
From e^(-6y) = 3, we can take the natural logarithm of both sides to solve for y:
-6y = ln(3)
y = -ln(3)/6.
From cos(6x) + i*sin(6x) = i, we can equate the real and imaginary parts:
cos(6x) = 0
sin(6x) = 1.
From cos(6x) = 0, we know that 6x = (2n + 1) * (π/2), where n is an integer.
Thus, x = (2n + 1) * (π/12).
Now we have the Cartesian form of z = x + yi, where x = (2n + 1) * (π/12) and y = -ln(3)/6.
To find the complex numbers that satisfy the equation, we can substitute different values of n into the expressions above and get the corresponding values of x and y. Each (x, y) pair represents a complex number in Cartesian form. Then, we can plot these complex numbers on the complex plane to visualize their positions.
I hope this explanation helps! If you have any further questions, feel free to ask.