The question is

http://www.webassign.net/www22/latex...d58f9d7ef1.gif

im not sure how to do this, i know that i=sqrt(-1) but im not sure how to compute this...

thank you

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- August 7th 2009, 09:53 PMacosta0809eulers
The question is

http://www.webassign.net/www22/latex...d58f9d7ef1.gif

im not sure how to do this, i know that i=sqrt(-1) but im not sure how to compute this...

thank you - August 7th 2009, 10:12 PMred_dog

- August 7th 2009, 10:26 PMFailure
You don't get a unique answer to this:

a) (Thinking) where

So, basically you have the two answers (Thinking) and (Thinking)

Therefore, wirting an equality sign seems rather problematic. Rather, we would want to say that the equation has two solutions, and .

b) Just use for the above two solutions . - August 7th 2009, 11:46 PMalunwComplex square root without trig
a)Almost every number has two distinct square roots. But conventionally denotes the positive solution to . So the usual convention for square roots of complex numbers is to say that is the square root with a positive real part. If x is real and negative then is the root with positive imaginary part. This means there is a branch cut along the negative real axis.

b) In fact you can find find square roots without resorting to trigonometry. The following pseudo-code, computes the square root (x,y) of the complex number (u,v)

Code:`if (u > 0) then`

x = sqrt((u+sqrt(u*u+v*v))/2)

y = v/(2*x)

else if (u < 0) then

y = sign(v)*sqrt((-u+sqrt(u*u+v*v))/2)

x = v/(2*y)

else

x = sqrt(abs(v)/2)

if (v > 0)

y = x

else

y = -x