Let Z be a complex number.whicj of the following is solution set of z^3-iz=0
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Don't see any "following", but expect to see something like ....
z^3-iz=0
z(z^2 - i) = 0
z = 0 or z = ±√ i
Now about that √ i
let a + bi = √ i
a^2 + 2abi + b^2 i^2 = i
a^2 - b^2 + 2abi = 0 + i
a^2 - b^2 = 0
a = ±b
and
2abi = i
2ab = 1
2a^2 = 1
a = ±1/√2 = ±√2 / 2
z=0 , z = ±√2 / 2 ± (√2 / 2) i = ±(√2/2)(1 + i)
To find the solution set of the equation z^3 - iz = 0, we can solve it by factoring.
Let's start by factoring out z from both terms:
z(z^2 - i) = 0
Now, set each factor equal to zero, and solve for z:
z = 0 (from the factor z = 0)
z^2 - i = 0
To solve the quadratic equation z^2 - i = 0, we can use the quadratic formula.
The quadratic formula states that for an equation of the form ax^2 + bx + c = 0, the solutions are given by:
x = (-b ± √(b^2 - 4ac)) / 2a
Applying this to our equation z^2 - i = 0, we have:
a = 1, b = 0, and c = -i
Plugging these values into the quadratic formula, we get:
z = [ -0 ± √(0^2 - 4(1)(-i)) ] / 2(1)
= [ ± √(0 - 4(-i)) ] / 2
= [ ± √(0 + 4i) ] / 2
= [ ± √(4i) ] / 2
Now, let's simplify √(4i). To find the square root of a complex number, we can express it in polar form and then take the square root.
4i can be expressed as 4 * (cos(π/2) + i*sin(π/2)) in polar form.
Now, taking the square root of 4 * (cos(π/2) + i*sin(π/2)), we get:
√(4i) = √(4) * (cos(π/4) + i*sin(π/4))
= 2 * (cos(π/4) + i*sin(π/4))
Substituting this back into the expression for z, we have:
z = [ ± 2 * (cos(π/4) + i*sin(π/4)) ] / 2
= ± (cos(π/4) + i*sin(π/4))
Finally, we can simplify cos(π/4) and sin(π/4) as follows:
cos(π/4) = (1/√2)
sin(π/4) = (1/√2)
Therefore, the solution set for z^3 - iz = 0 is:
z = 0, ± (1/√2 + i/√2)