Question

\[\sqrt{a+ib}\cdot\sqrt{\frac1{a+ib}}=\begin{cases} ~\\~\\~\end{cases}\]

Answer 1

first step should be square both sides

Answer 2

but what is the RHS? @mathslover

Answer 3

let RHS be x

Answer 4

how do you know there is only one solution?

Answer 5

oh k wait i have an easier way

Answer 6

it is one from your eyeballs

Answer 7

@satellite73
how do you know there is only one solution?

Answer 8

not sure what you mean

Answer 9

well my eyes see that if a=b=0 then the is no solution

Answer 10

\[\large{\frac{(a+ib)(a-ib)}{a^2+b^2}=x^2}\]
\[\large{\frac{a^2+b^2}{a^2+b^2}=x^2}\]
\[\large{1=x^2}\]
\[\large{x=1}\]

Answer 11

\[\sqrt{z}\times\frac{1}{\sqrt{z}}=\sqrt{\frac{z}{z}}=\sqrt{1}=1\]

Answer 12

that is your answer

Answer 13

right satellite :D

Answer 14

\[\sqrt{a+ib}\cdot\sqrt{\frac1{a+ib}}=\begin{cases} ~\\~\\~\end{cases}\]
Which is equal to \[\sqrt{\frac{ a+ib }{ a+ib }} = \sqrt{1} = \pm 1\]

Answer 15

\pm matters a lot

Answer 16

got it @UnkleRhaukus ?

Answer 17

x = \(\pm 1\)

Answer 18

A lot of lot.

Answer 19

unklerhaukus is typing a reply :)

Answer 20

\[\sqrt{a+ib}\cdot\sqrt{\frac1{a+ib}}=\begin{cases} 1&...\\0& a=b=0\\-1&...\end{cases}\]

Answer 21

depends on whether you think \(\sqrt{a}\) means the the positive root of \(a\) square root is a well defined function, or whether you think \(\sqrt{a}\) means any solution to \(x^2=a\)
i am of the former view, but that is just my opinion

Answer 22

really? what number is \(\frac{1}{0}\)?

Answer 23

whops,

Answer 24

\[\sqrt{a+ib}\cdot\sqrt{\frac1{a+ib}}=\begin{cases} 1&...\\ ? & a=b=0\\-1&...\end{cases}\]

Answer 25

It is still 0 if a=b=0.

Answer 26

yes but that is not possible

Answer 27

In fact no, it's still \pm 1 if a=b=0

Answer 28

\(\frac{0}{0}=1\) ?

Answer 29

I think the given problem needs something more info. :
\[\large{\textbf{IF} a \space \textbf{and} \space b\ne 0}\]

Answer 30

No... they cancel.

Answer 31

\[\frac{0}{0}=\frac{\cancel{0}}{\cancel{0}}=1\]

Answer 32

\[a \times \frac{ b }{ a }\] = b. We can all agree on that can't we?
Just because a can take the value zero doesn't mean that there is a different solution there.
The fact remains that the a's cancel.

Answer 33

i like cancelling zeros

Answer 34

0/0 = undefined

Answer 35

but even in the world of complex numbers, you cannot divide by 0

Answer 36

But you aren't dividing by 0, it has been cancelled.

Answer 37

implicit is writing \(\frac{1}{z}\) is that \(z\neq 0\)

Answer 38

Yes this should have been written in the question really.

Answer 39

no it is in there already

Answer 40

guys let unkle rhaukus think also :D

Answer 41

ok from now on Z≠0

Answer 42

i guess i ldepends on how you view this symbol
\[\sqrt{a}\]

Answer 43

yeah i have been thinking about it today http://openstudy.com/users/unklerhaukus#/updates/502f3359e4b0ac2883165a00

Answer 44

to some, myself included, a real number has two "square roots"
the two square roots of \(a\) are \(\sqrt{a}\) and \(-\sqrt{a}\) meaning the symbol \(\sqrt{a}\) means the "positive square root" i.e. the positive numbers whose square is \(a\)

Answer 45

\[\sqrt{a+ib}\cdot\sqrt{\frac{1}{a+ib}}\]\[=\sqrt{a+ib}\cdot\sqrt{\frac{1}{a+ib}\times\frac{a-ib}{a-ib}}\]\[=\sqrt{a+ib}\cdot\sqrt{\frac{a-ib}{a^2+b^2}}\]

Answer 46

we can try a little experiment and see what the wolf gods think
http://www.wolframalpha.com/input/?i=sqrt%282%2B3i%29*1%2F%28sqrt%282%2B3i%29%29

Answer 47

dont you mean
http://www.wolframalpha.com/input/?i=sqrt%282%2B3i%29*sqrt%281%2F%282%2B3i%29%29,

different result

Answer 48

that is what i did @UnkleRhaukus

Answer 49

Computation timed out. @satellite73

Answer 50

mine gave me 1 instantly

Answer 51

i used your link and got one as well
did it time out for you?

Answer 52

well it does say one 1,
it also say Computation timed out.

which explains the missing solutions ,


I have been trying to prove (11)
http://mathworld.wolfram.com/SquareRoot.html

Answer 53

http://www.wolframalpha.com/input/?i=sqrt%28a%2Bbi%29*sqrt%281%2F%28a%2Bbi%29%29
\(\downarrow\)
http://www.wolframalpha.com/input/?i=e%5E%28i+%CF%80+floor%281%2F2%2B%28arg%28a%2Bi+b%29%29%2F%282+%CF%80%29%29%29&lk=1&a=ClashPrefs_*Math-

im not understanding

Answer 54

it is graphing it for you in two dimensions, as if you are in the \((a,b)\) plane
but you notice that it is 1 straight across

Answer 55

yeah i see that bit of the graph but how did can i arrive at that

Answer 56

ok look all the way at the bottom
you will get a better answer than i provided

Answer 57

the scary looking ∑ sums?

Answer 58

interesting, wolfram is contradicting itself
here check this out
http://www.wolframalpha.com/input/?i=sqrt%28-2-3i%29*sqrt%281%2F%28-2-3i%29%29%2C

Answer 59

hmm machine problem , ,

how do they get the argument thing \(\text{Arg(a+ib)}\)

Answer 60

\[ = \left\{\begin{array}{rcc}
-1 & \text{if} & arg(a+bi)\geq \pi \\
1& \text{othewise}
\end{array}
\right.
\]

Answer 61

is that from a picture ?