Question

x = sqrt(1+i) -sqrt(1-i) for x.

Answer 1

Wolfram say s the square root of i-1 and i+1 are the same...

Answer 2

so zero !!! O_o I'm confused

Answer 3

Can't get the link to post correctly....

Answer 4

How about Polar Form ?

Answer 5

Lets say z = 1 -i

Answer 6

Put

sqrt(1+i) - sqrt(1-i)

in Wolfram....

Answer 7

i sqrt (2(sqrt2-1))....

Answer 8

if i do that in polar i get just 1 -i
\[\sqrt{2}e^{i \frac{3\pi}{4}}\]

Answer 9

then for 1+i it becomes same but with pi/4

Answer 10

ANd why hasn't sqrt(1-i) or (1+i) actually got 2 roots not 1?

Answer 11

I'm confused::¿

Answer 12

http://www.wolframalpha.com/input/?i=sqrt%281%2Bi%29+-+sqrt%281-i%29

Answer 13

Me too !

Answer 14

Ur having same trouble with link as me, it's not right...

Answer 15

Decimal Approximation .90.. something

Answer 16

That what Wolfram is showing

Answer 17

They must be "choosing" a square root but what basis, u can't say the "positive" one..

Answer 18

http://www.wolframalpha.com/input/?i=simplify+sqrt%281%2Bi%29+-+sqrt%281-i%29

Answer 19

Grr.. can't get the links to come up right....

Answer 20

Put simplify sqrt etc in Wolfram

Answer 21

Ok

For Polar it should look like this

\[x =\sqrt{\sqrt{2}e^{i \frac{\pi }{4}}} - \sqrt{\sqrt{2}e^{i \frac{3 \pi}{4}}} \]

Answer 22

Now u have the eîpi = 1 thingy....

Answer 23

So you split the pi/4 into 1 and eî whatever.

Answer 24

e^(1/4)

Answer 25

Yeah then you take sqrt{sqrt2 }common and it should become

\[2^{1/4} \times ( e^{i \frac{\pi}{8}} - e^{i \frac{3 \pi }{8}})\]

Answer 26

sqrt(sqrt2 e^(1/4))....lol

Answer 27

then when we open it into e^i(theta ) into cos theta + i sin theta

Answer 28

lol

Answer 29

I gave u medal for trying to help....:-)

Answer 30

>What a mess!

Answer 31

thanks

Answer 32

with it should become \[2 \times i \times 2^{1/4} sin\frac{\pi}{8}\]

Answer 33

Now we have to solve this

Answer 34

Yes they have like that as alternate form in Wolfram....

Answer 35

But still, they are choosing a root...

Answer 36

yeah

Answer 37

And they are choosing the same root for both terms!!!!

Answer 38

Maybe this has something to do with it..?

http://en.wikipedia.org/wiki/Branch_cut#Branch_cuts

Answer 39

I am so confused now ..

Answer 40

Let's leave it and come back to it later....?

Answer 41

Yeah maybe others Like joemath and Zarkon can clear this ..they are really good ..I would have to read about it I'l come back later after reading enough then try this ..

Answer 42

Latest question...what is the last number of the pi? :-)

Let's go solve that one...

Answer 43

\[x ^{2}=1+i+1−i+2(1−i ^{2})\] should be
\[x ^{2}=1+i+1−i+2\sqrt{(1−i ^{2})}\]

You will eventually get x= sqrt(stuff), so x will be plus or minus this expression.

As for the \[\sqrt{i+1}\]
It's most easily done in polar coordinates. e.g.
\[1+i = \sqrt{2}e ^{\frac{i \pi}{4}}\]

and its square root is
\[(1+i)^{\frac{1}{2}}= \pm2^{\frac{1}{4}} e ^{\frac{i \pi}{8}}\]

Of course you can re-write the negative root as\[2^{\frac{1}{4}}e ^{\frac{i \pi}{8}} e ^{i\pi} = 2^{\frac{1}{4}}e ^{\frac{i 9 \pi}{8}}\]

Answer 44

Satellite, u got a take on this one?

Answer 45

oops, that should be a - 2 sqrt(1- i*i) in the starting equation.
i.e.
\[x^{2}=1+i+1−i−2\sqrt{(1−i^{2})}\]
which simplifies to
\[x^{2}=2−2\sqrt{2}\]
and
\[x= \pm\sqrt{2(1−\sqrt{2})}\]
Now, because the 1-sqrt(2) is negative, we can re-write this as
\[x= \pm i\sqrt{2(\sqrt{2}-1)}\]

Tough typing this in without the random mistake!

Answer 46

I think I have id'd the problem with this now.

It depends on whether u allow x to take complex values.

The root 2 thingy is if x can take complex and the other is if x is real.

Answer 47

And Wolfram seem to arbitrarily choose the expression with positive i as a kind of "positive" root and then take the other for the conjugate.

Answer 48

Here's what I think is going on...but I haven't really checked...
First, we are working on the complex plane.
Now let's take the nth root of a number.
The number in polar form is magnitude * exp(i theta)
its first root is magnitude^(1/n) * exp(i theta/n)
Of course there are n roots...
we can always add 2pi to the argument (theta), so the next root
is at magnitude^(1/n) * exp( i theta/n + 2*pi/n)
ok let me ignore the magnitude..it's always the same...
the next root is at exp(i theta/n + 4*pi/n)
and so on until we have gone around the circle
notice that if the number is positive real we end up at exp(i*0) and exp(ipi) (i'm ignoring magnitude)
To summarize wolfram is just picking the 1st root at exp(i theta/n)

Answer 49

*notice that if the number is positive real we end up at exp(i*0) and exp(i pi)
I mean for the square root (n=2)

Answer 50

I am going to go back to my geometrical interpretation of i and forget all about complex analysis...:-)

Answer 51

I take it my explanation was not clear. Here's another try, with pictures!

Answer 52

Oh wow....interesting