Complex number help pls?

Question

find the roots of the equation

z^2 - z + 1 =0, giving answers in the form x +iy where x and y are real

Thanks

Answer 1

z = [ 1 ± √(1 - 4) ] / 2
z = [1 ± √ (- 3 )] / 2
z = [1 ± √ (3 i ²) ] / 2
z = [1 ± i √3 ] / 2

Answer 2

The determinant is (-1)^2 - 4, i.e. -3. Hence the solution becomes

(-(-1)±√-3)/2 = 1/2 ± i(√3)/2

I only used the quadratic formula like you did back in grade school or high school for quadratic equations :)

Answer 3

Hint:

This is a Quadratic Equation.

"nisss a" Has given you the answer, all you need to do is put the paramiters of you problem, in the equation and the result.

Root means solution, "i" is nothing but squar root of -1,

Any thing with i is a complex number, the teacher is is asking (loking) for both real and the coplex solutions of this problem.

Answer 4

Z2-Z+1 = 0
For a quadratic equation
Roots Z1 & Z2 will be given as
Z1, Z2 = {-b ± √(b2 – 4ac)}/2
Hence
Z1 = {1 + √ -3}/2
or Z1 = 1/2 + √3 i/2
and Z2 = 1/2 - √3 i/2

Answer 5

0.5 + i*0.5*√3 and 0.5 - i*0.5*√3
Use the quadratic formula for a*z^2 + b*z + c which gives the roots

(0.5/a)*(-b ± √[b^2 - 4*a*c])

here a = 1, b = -1, c = 1

Answer 6

its a qudratic equation in z^2 so use the formula:
for an eq:ax+by+c=0
x=-b+or-squarerootof(b^2-4ac)/2a the2a is divided where by th entire equation

Answer 7

z^3+1=0, z different than 1

z^3=-1= cos pi+ i sin pi

z=cos k pi/3 + i sin k pi/3, k from 1 to 2
or
z= cos pi/3 + i sin pi/3 and
z= cos 2pi/3 + i sin 2pi/3

Answer 8

the answer is

z (+/-) (z^2 - 4)/2

Where z is your number

z + (z^2 - 4)/2 and z - (z^2 - 4)/2