Question

MIT 6.189 Exercise 2.5

Answer 1

I didn't have too much trouble getting the first part of 2.5 done (see how I did it: https://gist.github.com/TomDeBeauchamp/5864128). I'd like to go for the optional second half, though, but I realize, I don't really remember complex numbers. I've looked at a few web pages on using the quadratic formula with complex numbers, but nothing really great.

Anyone have any suggestions for a good refresher tutorial on complex numbers, especially as regards the quadratic formula?

Answer 2

http://www.purplemath.com/modules/complex.htm

Answer 3

OH NOES, 404.

We seem to have missed the gist of that gist you were looking for.

Answer 4

Hmm? There is no 404. Just need to know how to read the link right:

https://gist.github.com/TomDeBeauchamp/5864128

Always put a space after a link and before punctuation, or the punctuation will corrupt the auto link generator.

Answer 5

http://pastebin.com/DVQVR7nW

Answer 6

Awesome. Thank you both for your help!

Answer 7

Since this question is closed, it is doubtful that anyone will find this code...
Here is how I finally settled on handling the complex roots, there is an 'import unicodedata' at the top. Python 3.x

if (x < 0):
print("Complex root")
x = -1*x
if (math.sqrt(x)%1 == 0):
r1 = -b/(2*a) + math.sqrt(x)/(2*a)*1j
r2 = -b/(2*a) - math.sqrt(x)/(2*a)*1j
print ("Root 1 =", r1)
print ("Root 2 =", r2)
else:
# part1 + sqrt(part2)j/part3
part1 = -b/(2*a)
part2 = x*1j
part3 = 2*a

print ("Root 1 = ", part1, " + ", unicodedata.lookup('square root')+"("+ str(part2) + ")/" + str(part3) )
print ("Root 2 = ", part1, " - ", unicodedata.lookup('square root')+"("+ str(part2) + ")/" + str(part3) )

Answer 8

I'm a little new... Is there a way to re-open questions?

Answer 9

Not that I am aware of. There is a "Closed Questions" tab that you can use to review those that have been finalized, but I think starting a new thread is your only option to revisit old topics.

Answer 10

whoops... I'll keep that in mind going forward.

btw, this bit was the life-saver:

r1 = -b/(2*a) + math.sqrt(x)/(2*a)*1j
r2 = -b/(2*a) - math.sqrt(x)/(2*a)*1j

Answer 11

Glad to hear it.

Answer 12

Not sure if there is a policy about adding to closed questions, especially ones 7 months old but here's what I did.

I solved this by using import cmath at the top of the code which allows you to use cmath.sqrt for negative numbers:
'if discriminant < 0
complex1 = (-b + cmath.sqrt(discriminant)) / ( 2 * a)
complex2 = (-b - cmath.sqrt(discriminant)) / ( 2 * a)
return complex1, complex2

#testcase:
print quadratic_root(1, 2, 3)
results in ((-1+1.4142135623730951j), (-1-1.4142135623730951j))