find the value of x and y that make each equation true -4+(y)^i=-12x-i+8
please show the steps thanks
the real and imaginary parts must be equal, so
-4 = -12x+8
y = -1
x = 1
y = -1
Well, solving this equation might not be as straightforward as it seems. You see, there's a sneaky little "i" in there, which stands for the imaginary unit (√-1). So things are about to get a little imaginary up in here!
Let's break it down step by step:
Step 1: Group the real and imaginary parts of the equation.
The real parts are -4 and -12x + 8. The imaginary parts are yi and -i. So we can rewrite the equation as:
-4 = -12x + 8
And
yi = -i
Step 2: Solve for x
For the real part, -4 = -12x + 8, we can subtract 8 from both sides to get:
-12 = -12x
Divide both sides by -12:
x = 1
Step 3: Solve for y
For the imaginary part, we have yi = -i. This equation tells us that the real part of yi is 0 and the imaginary parts equal each other, so:
y = -1
So the solution to the equation is x = 1 and y = -1. And that's no imaginary tale!
To find the values of x and y that make the equation true, we need to separate the real and imaginary parts of the equation and solve them separately.
Given equation: -4 + y^i = -12x - i + 8
Separating the real and imaginary parts:
Real part: -4 = -12x + 8
Imaginary part: y^i = -i
Solving the real part:
-4 = -12x + 8
-12x = -12
x = -12 / -12
x = 1
Now, let's solve the imaginary part:
To get rid of the imaginary unit "i" on both sides of the equation, we'll raise e to the power of both sides.
e^(y^i) = e^(-i)
By Euler's formula, we can express e^(ix) as cos(x) + i sin(x):
e^(y^i) = cos(-1) + i sin(-1)
Now we'll compare the real and imaginary parts separately:
Real part:
e^(y^i) * cos(1) = cos(-1)
Imaginary part:
e^(y^i) * sin(1) = sin(-1)
Simplifying the real part:
cos(1) = cos(-1)
Since the cosine function is an even function, cos(-1) = cos(1).
Simplifying the imaginary part:
sin(1) = -sin(-1)
Since the sine function is an odd function, sin(-1) = -sin(1).
Therefore, the values of x and y that make the equation true are:
x = 1
y = 1
To find the values of x and y that make the equation true, we need to simplify both sides of the equation and compare the corresponding real and imaginary parts.
Let's break down the given equation step by step:
-4 + (y)^i = -12x - i + 8
To simplify, we separate the real and imaginary parts on both sides using Euler's formula:
-4 + y * (cos(i * ln(y)) + i * sin(i * ln(y))) = -12x - i + 8
Comparing the real parts:
-4 + y * cos(i * ln(y)) = -12x + 8
Comparing the imaginary parts:
y * sin(i * ln(y)) = -1
Now we have two equations:
1) -4 + y * cos(i * ln(y)) = -12x + 8
2) y * sin(i * ln(y)) = -1
To solve equation 2, we need to consider the values of y that satisfy it. The equation -1 = y * sin(i * ln(y)) has an infinite number of solutions, so we need to find the values of y that yield -1 when multiplied by sin(i * ln(y)).
Next, let's address equation 1: -4 + y * cos(i * ln(y)) = -12x + 8
Since y can take on any value satisfying equation 2, let's solve equation 1 for x:
y * cos(i * ln(y)) = -12x + 12
Divide both sides by -12:
(x - 1) = -y * cos(i * ln(y)) / 12
Now, on the right side, we can substitute the value of y * cos(i * ln(y)) / 12 using the equation (-1) * sin(i * ln(y)):
(x - 1) = -(-1) * sin(i * ln(y))
(x - 1) = sin(i * ln(y))
Take the inverse sine of both sides:
asin(x - 1) = i * ln(y)
Now, we have an expression involving complex numbers. To convert it into exponential form:
e^(i * asin(x - 1)) = e^(i * (i * ln(y)))
Using Euler's formula, e^(i * asin(x - 1)) can be expressed as:
cos(asin(x - 1)) + i * sin(asin(x - 1))
Similarly, e^(i * (i * ln(y))) can be written as:
cos(i * ln(y)) + i * sin(i * ln(y))
Comparing the real parts:
cos(asin(x - 1)) = cos(i * ln(y))
Comparing the imaginary parts:
sin(asin(x - 1)) = sin(i * ln(y))
Remember from equation 2, we had:
y * sin(i * ln(y)) = -1
This means that sin(i * ln(y)) = -1/y.
Substituting this into our equation:
sin(asin(x - 1)) = -1/y
Since sin(asin(x - 1)) = x - 1, we get:
x - 1 = -1/y
Solving this equation for y:
y = -1/(x - 1)
Now, we have expressions for x and y:
x = (y - 1) and y = -1/(x - 1)
We can substitute one equation into the other to obtain the final solution.