complex numbers / Quadratic equation: (1-i)*z^2+(1+3i)z+(-4+

[DarkSun]

I have the following problem:
Solve the quadratic equation: (1-i)*z^2+(1+3i)z+(-4+2i)=0

So what I tried is: a=1-i, b=1+3i, c=-4+2i
Then plugged them inside the quadric formula, and eventually got:
z1,2 = -(1+3i)+-sqrt( (1+3i)^2 -4 (1-i)(-4+2i) ) / 2(1-i)
...
= (-1-3i+-sqrt(-18i)) / 2-2i

now I guess this sqrt(-18i) should be broken up somehow...so:
1) sqrt(-18i)=sqrt(18)*sqrt(-i)
sqrt(18) = sqrt(9*2)=3*sqrt(2)
2) sqrt(-i) = sqrt( - sqrt(-1) ) = ...?
what can I break this down...? or can't I?

Thanks.

 

[mmm4444bot]

I think there are at least three different methods for calculating the square root of a non-Real Complex number. (Use Google.)

sqrt(-i) = sqrt(2)/2 * (1 - i)

I get 2 - i and -1 - i for the roots of the given quadratic equation. And you?

 

[DarkSun]

Re: complex numbers / Quadratic equation

It's not calculating the root that confuses me, but instead this algebric step -
sqrt(-i) = sqrt(2)/2 * (1 - i)
Why is that so? I know - sqrt(-i) = sqrt( - sqrt(-1) ) but how do you get from this to sqrt(2)/2 * (1 - i) ?

 

[pka]

Re: complex numbers / Quadratic equation

Why is that so? I know - sqrt(-i) = sqrt( - sqrt(-1) ) but how do you get from this to sqrt(2)/2 * (1 - i) ?

\(\displaystyle - i = e^{\frac{{ - i\pi }}{2}} \, \Rightarrow \,\left( { - i} \right)^{\frac{1}{2}} = e^{\frac{{ - i(\pi + k\pi) }}{4}} ,\;k = 0,1\)
Recall that:\(\displaystyle \cos \left( {\frac{{ - \pi }}{4}} \right) = \frac{{\sqrt 2 }}{2}\;\& \,\sin \left( {\frac{{ - \pi }}{4}} \right) = \frac{{ - \sqrt 2 }}{2}\).
Therefore \(\displaystyle e^{\frac{{ - i\pi }}{4}} = \frac{{\sqrt 2 }}{2} - \frac{{\sqrt 2 }}{2}i = \frac{{\sqrt 2 }}{2}\left( {1 - i} \right)\)

 

[mmm4444bot]

It's not calculating the root that confuses me, but instead this algebric step -
sqrt(-i) = sqrt(2)/2 * (1 - i)

Well, my friend, if you're confused about how to calculate this root, then it makes no sense for you to post "it's not calculating the root that confuses me".

I told you plainly that there are multiple methods for calculating the root of -i.

I told you to use Google; you can find thousands of explanations and examples on these methods.

Are you lazy?

I am. I'm too lazy to reproduce explanations and examples on these methods because others have typed them all up for us to read -- THOUSANDS of times over.

 

[DarkSun]

Sorry, you were right,
I was talking nonsense before, I understand how it's done now.