Finding Solutions for a Quadratic Equation with Complex Numbers

[ihumayun]

Homework Statement

Solve the following equation for Z, find all solutions.

z² -2z + i = 0

Homework Equations

[-b(+/-) sqrt(b²-4ac)]/2a

The Attempt at a Solution

Using the equation above,

z = [2 (+/-) sqrt( (-2)² - 4 (1) (i)) ]/ 2(1)

=[2 (+/-) sqrt ( 4 - 4i)]/2

= [2 (+/-) sqrt ( (4) (1-i)]/2

= [2 (+/-) sqrt(4) sqrt(1-i)]/2

= [2 (+/-) 2 sqrt(1-i)]/2

= 1 (+/-) sqrt (1-i)

which means the roots are 1+ sqrt(1-i) and 1- sqrt(1-i). This is the answer that I have entered into my online assignment, but it is being marked as incorrect. Do I have an error in my calculations, or can the answer can be simplified further?

 

[njama]

Original problem:

$$z^2-2z+i=0$$

Check of your solution:

$$z_{1,2}=\frac{2 \pm \sqrt{4-4i}}{2}$$

$$z_{1,2}=\frac{2 \pm 2\sqrt{1-i}}{2}$$

$$z_{1,2}=1 \pm \sqrt{1-i}$$

Another approach:

$$(z-1)^2-1+i=0$$

$$(z-1)^2=1-i$$

$$z-1=\pm \sqrt{1-i}$$

$$z = 1 \pm \sqrt{1-i}$$

 

[ihumayun]

So I have it right then. I guess the site is being picky with the answer format or something. Thank you!