Question

If SqRt(i) = a + ib, find a and b.

Answer 1

\[\sqrt{i}=a+ib\]

Answer 2

\(\sqrt{i}\) - ib = a

Answer 3

it cant be that simple... there has to be something im missing here.

Answer 4

now find b

Answer 5

solving a and b that way, is basic algebra, this unit is over complex numbers, i dont think it would be that easy

Answer 6

anyone know what to do?

Answer 7

what about

(\(\sqrt{i}\))^2 = (a + ib)^2

Answer 8

^^That would be my preferred method. Since \(\sqrt{i}^2=i\), this way would be fairly straightforward.

Answer 9

yes, but that only gives a trinomial equaling -1

Answer 10

okay even if I squared both sides.
\[i=a ^{2}-b ^{2}+2iab\]

Answer 11

how does that help me

Answer 12

the one thing I can think of is, the real part may only go to 0
the imaginary part "bi" can then be simplified

Answer 13

By setting imaginary parts equal, and real parts equal, you now have two equations for two unknowns.\[a^2-b^2=0\]\[2ab=1\]

(also, if you've learned about polar form, there is another method you could try)

Answer 14

I type it into WA and it tells me
\[a=\sqrt[4]{-1}-ib=(\sin 45 + isin 45) - ib\]

Answer 15

i honestly dont remember this to well, its been a good 12 years since i was in calc.

Answer 16

i know basics, but not properties or identities of it

Answer 17

Let's try using the polar notation method. Since \(i\) is a distance of 1 from the origin, and it lies on the vertical axis (which is at an angle of \(\pi/2\)), we can write\[i=\cos(\pi/2)+i\sin(\pi/2)=e^{i\pi/2}.\]Now, we can take the square root of \(e^{i\pi/2}\). \[\sqrt{e^{i\pi/2}}=\left(e^{i\pi/2}\right)^{1/2}=e^{i\pi/4}.\]Just convert back into the standard notation, and you've got it.

Answer 18

okay. but all you did was do the left side of the equation right...
so e^ipi/4 = a + bi.... i so are you solving a, or what are u doing

Answer 19

You are solving for \(a\) and \(b\) now. \(e^{i\pi/4}\) defines a unique complex number, so it is possible to solve for \(a,b\).

Answer 20

Do you know how to convert back and forth between these two notations?

Answer 21

no i do not remember, im sure once you start though it will probably come back

Answer 22

but im assuming its the sin 45 or something since thats what WA said

Answer 23

cuz pi/4 is 45 degrees right

Answer 24

So the notation is just in the form \[\large re^{i\theta}\]where \(r\) is the distance from the origin, and \(\theta\) is the angle from the positive real axis. What I wrote up there, has \(r=1\) and \(\theta=\pi/4\). As it happens, \(\pi/4=45^{\circ}\) when you convert to degree.

Answer 25

okay, so that is where the sin 45 + isin 45 comes from?

Answer 26

The other form, that's usually an intermediate step in switching between forms, is written as \[r(\cos(\theta)+i\sin(\theta))\]with \(r,\theta\) being the same as above. So we have \[\cos(\pi/4)+i\sin(\pi/4).\]These can easily be evaluated by inspection, WA, or any kind of calculator.

Answer 27

so its saying a = cos (pi/4) and b is sin(pi /4)

Answer 28

Exactly. And those can be evaluated to get a value.

Answer 29

dont you think it would be better to leave it like that or to put it as the decimal .70711

Answer 30

I would put it as \(\sqrt2/2\).

Answer 31

ohh okay.. a=b=sq2/2... thank you very much.... would u be able to help me with one more real quick

Answer 32

If it's quick. I have about 5-10 minutes before I have to go.

Answer 33

\[\frac{ 1 }{ 2-i }=a+ib\]

Answer 34

You want to multiply by\[\frac{2+i}{2+i}\]This is equivalent to multiplying by 1, but it'll get rid of the imaginary numbers in the denominator.

Answer 35

okay, so 2/5 + 1/5i
then you convert to 2/5 + (1/5)e^ipi/2

Answer 36

Don't convert to anything with \(e\). Just leave it like \[2/5+i/5\]since that's the form you want it in.

Answer 37

ohhh, your right, so a is 2/5, and b is 1/5... fantastic. wish you didnt have to leave... thanks for the help.

Answer 38

You're welcome.