(z − 1)^2 + 4 = 0
(z-1)^2 = -4
z-1 = ±2i
z = 1±2i
if solving for z then you can foil out (z-1)^2 and add the constant to 4
you should have an equation in the form of ax^2+bx+c
Set the equation equal to 0 and solve for z. I would use the quadratic equation.
Your final answer with have imaginary numbers.
To solve the equation (z − 1)^2 + 4 = 0, we can follow these steps:
Step 1: Expand the equation
(z − 1)^2 + 4 = 0
z^2 - 2z + 1 + 4 = 0
z^2 - 2z + 5 = 0
Step 2: Recognize the quadratic form
The equation z^2 - 2z + 5 = 0 is a quadratic equation in the form of ax^2 + bx + c = 0, where a = 1, b = -2, and c = 5.
Step 3: Use the quadratic formula
The quadratic formula is given by x = (-b ± √(b^2 - 4ac)) / (2a).
Plugging in the values a = 1, b = -2, and c = 5 into the quadratic formula:
z = (-(-2) ± √((-2)^2 - 4(1)(5))) / (2(1))
z = (2 ± √(4 - 20)) / 2
z = (2 ± √(-16)) / 2
Step 4: Calculate the square root of -16
Since the square root of -16 is not a real number, we know that this equation has no solution in the set of real numbers.
Step 5: Conclusion
The equation (z − 1)^2 + 4 = 0 has no solution in the set of real numbers.