Find all solutions and leave answers in trigonometric form.
x^2 - i = 0
Please help! I have no idea how to solve this!
x^2 = i
x = ±√i
Now, i = (1,π/2)
so x = ±(1,π/4)
or, (1,π/4), (1,5π/4)
let x = a + bi , where a and b are real numbers
x^2 = a^2 + 2abi + b^2i^2, but i^2 = -1
= a^2 - b^2 + 2abi
= i <----- that was given
by comparison, a^2 - b^2 = 0
a^2 = b^2
a = ± b
and 2abi = i
2ab = 1
if a= +b, 2b^2 = 1
b = ± 1/√2 , and a = ± 1/√2 or ± √2/2
if a = -b, 2b(-b) = 1
b^2 = -1/2, no solution for real values of b
so x = a + bi
= (√2/2 + √2/2i) OR (-√2/2 + √2/2i)
= 1(cos45° + i cos45°) or 1(-cos225° + i sin225°)
= ± cosπ/4 + i sinπ/4
check one of them
x = √2/2 + √2/2i
x^2 = (√2/2 + √2/2i)^2
= 1/2 + 2(1/2) i + (1/2) i^2
= 1/2 + i - 1/2
= i
or, using De Moivre's theorem
x^2 = 1^2 (cos 2(45°) + i sin 2(45°) )
= cos90 + isin90
= 0 + 1i = i
To solve the equation x^2 - i = 0, we can use the quadratic formula. However, instead of using real numbers, we will use complex numbers in the form a + bi.
The quadratic formula is given by:
x = (-b ± √(b^2 - 4ac)) / 2a
In this case, the equation is x^2 - i = 0, so we have:
a = 1
b = 0
c = -i
Plugging these values into the quadratic formula, we get:
x = (-0 ± √((0)^2 - 4(1)(-i))) / 2(1)
Simplifying:
x = ± √(4i) / 2
x = ± √2i / 2
To simplify further, we can rationalize the denominator by multiplying the numerator and denominator by the conjugate of the denominator (√2i / 2):
x = ± (√2i / 2) * (√2i / √2i)
x = ± (√2i * √2i) / (2 * √2i)
Applying the property √ab = √a * √b:
x = ± (−√2) / (2 * √2) * i
x = ± (√2 / 2) * i
Therefore, the solutions in trigonometric form are:
x = ± (√2 / 2) * i
To solve the equation x^2 - i = 0, we want to find the values of x that make the equation true.
First, let's rearrange the equation:
x^2 = i
To solve for x, we can take the square root of both sides of the equation. However, since we need to express the solutions in trigonometric form, we will use the polar form of complex numbers.
In polar coordinates, a complex number z = r(cosθ + isinθ), where r represents the magnitude, and θ represents the angle from the positive real axis in the counterclockwise direction.
Now, let's convert i to its polar form.
To find the magnitude (r):
|r| = sqrt(1^2 + 0^2) = 1
To find the angle (θ):
θ = arctan(0/1) = 0
Thus, i can be represented as i = 1(cos0 + isin0).
Now, let's find the square root of i:
sqrt(i) = sqrt(1(cos0 + isin0))
To find the square root of a complex number in polar form, we take the square root of the magnitude and divide the angle by 2.
|r|^(1/2) = 1^(1/2) = 1
θ/2 = 0/2 = 0
Therefore, sqrt(i) = 1(cos0/2 + isin0/2)
Simplifying further, we have:
sqrt(i) = 1(cos0 + isin0)^(1/2) = 1(cos0/2 + isin0/2)
Now, we have the square root of i in trigonometric form.
To find the solutions for x, we can use euler's formula which states that e^(ix) = cos(x) + isin(x).
So, if we equate sqrt(i) to sqrt(1) * sqrt(cos0/2 + isin0/2), we can express the solutions as:
x = ±1^(1/2) * (cos (0/2) + isin (0/2))
Simplifying further, we have:
x = ±1^(1/2) * (cos 0 + isin 0)
Now, let's evaluate the expressions for the positive and negative square roots of 1:
For the positive square root (x1):
x1 = sqrt(1) * (cos 0 + isin 0) = 1 * (1 + i0) = 1
For the negative square root (x2):
x2 = - sqrt(1) * (cos 0 + isin 0) = -1 * (1 + i0) = -1
Therefore, the solutions to the equation x^2 - i = 0, expressed in trigonometric form, are x = 1 and x = -1.