Need help with 2 questions from you guys (1 Viewer)

[currysauce]

Complex Numbers

solve the following quadratic equation

ix^2 - 2( i + 1 )x + 10 = 0

i can't get it after i find x and iy for z

anyway


2.

1-2i is one root of x^2 - (3+i)x + k = 0. Find k and the other root of the equation.

 

[Estel]

1. Standard quadratic equation... post your working if you still have problems.
2. (1-2i)+z = (3+i)
so the other root is (2+3i), and using product of roots you can find k.

 

[currysauce]

thankyou good sir

 

[Slidey]

ix^2 - 2( i + 1 )x + 10 = 0

i(x-[i+1])^2 + 10 -i[i+1] = 0

i(x-[i+1])^2 = -10 +i[i+1] = -10 -1 + i

(x-[i+1])^2 = (-11 + i)/i = 1+11i

x-i+1=+sqrt(1+11i)

Now, let 1+11i=(x+yi)^2
1+11i=x^2-y^2 + 2xyi
1=x^2-y^2
11=2xy
y=11/(2x)
x^2-121/(4x^2)=1
x^4-x^2-121/4=0

x^2=w
x^2=[1+sqrt(122)]/2
x=+sqrt{[1+sqrt(122)]/2}

x=11/(2y)
1=(121/4)/y^2-y^2
y^2=121/4-y^4
y^4+y^2-121/4=0

y^2=[-1+sqrt(122)]/2
y=+sqrt{[-1+sqrt(122)]/2}

so sqrt(1+11i)=
+sqrt{[1+sqrt(122)]/2} +sqrt{[-1+sqrt(122)].i/2}
=
+{sqrt[1+sqrt(122)]+sqrt[-1+sqrt(122)].i}/sqrt2

Now back to the root:

x-i+1=+sqrt(1+11i)
x-i+1=+{sqrt[1+sqrt(122)]+sqrt[-1+sqrt(122)].i}/sqrt2
x=-1+{sqrt[1+sqrt(122)]+sqrt{[-1+sqrt(122)].i}/sqrt2 +i

Note the two roots are NOT complex conjugates of each other. This is because the coefficients of the quadratic were complex themselves.

I'm bored.