What is 2^2i?

Question

Also what is the square root of i and the square root of -i.

Answer 1

√i = ½√2 + i·½√2

√-i = -½√2 + i·½√2

Answer 2

Many of the answers posted so far are incomplete
and others are wrong.
First 2^(2i) = 4^i = e^(i log 4) = cos( log 4) + i sin(log 4).
Next, every complex number has 2 square roots.
So let's find both square roots of i.
First let's write i in polar form: i = 1/_π/2.
To get the square root of r/_θ, you
take √r and divide θ by 2.
So a square root of i is 1/_π/4 = (√2 + √2i)/2
The other square root is 1/_(π/4 + π) = (-√2 -√2i)/2.
To find the square roots of -i, use the fact that
-i = 1/_3π/2.
So √-i = 1/_3π/4 or 1/_7π/4 = (-√2 + √2i)/2
or (√2 -√2i)/2.
Here's another way to find the square roots of i and -i:
The square roots of i satisfy x²-i = 0
and the square roots of -i satisfy x²+i = 0.
Let's multiply these 2 equations together to get
x^4+1 = 0.
But this factors into 2 real quadratics:
(x² + √2x +1)(x² - √2x +1) = 0.
Now solve these 2 quadratics to get the 4
solutions above. By squaring, you can find
the ones whose square is i and those whose
square is -i.

Answer 3

Euler's Equation tells us...
e^(xi) = cos(x) + sin(x) i

1)
2^(2i) = 4^i = e(ln(4)*i) , so x=ln(4)

2^(2i) = cos(ln(4)) + sin(ln(4))i


2)
First put i in the e^(xi) form.
Is there a value for x that makes sin(x)=1 and cos(x)=0? Sure, pi/2.
i = cos(pi/2) + sin(pi/2)i = e^(i pi/2)
The square root of that is e^(i pi/4).
e^(i pi/4) = cos(pi/4)+sin(pi/4)i
= (1+i)sqrt(2)/2

Rearrange that however you want.

Let's check that:
[(1+i)sqrt(2)/2] ^ 2 = (1+i)^2 / 2
= (1 + 2i + i^2) / 2
= (1 + 2i - 1) / 2
= 2i/2
= i Perfect!

I'll let you figure out the square root of -i.

Answer 4

If you mean 2^(2i) instead of 2^2 * i = 4i:

2^(2i) = e^(2i ln 2) = cos (2 ln 2) + i sin (2 ln 2)

Also, in addition:

-sqrt(2)/2 - sqrt(2)/2 i and sqrt(2)/2 - sqrt(2)/2 i are the other square roots of i and -i, respectively. (In addition to the roots CogitoErgoCogitoSum gave)

edit: Alright... sue me. 2 ln 2 = ln 2^2 = ln 4.

Answer 5

I don't think 2^2i can be simplified any further

And you want x such that x^2=i

x=(-1)^1/4 (whatever that turns out to be

-i = i^3 = -1*i so it would be -(-1)^0.25

Answer 6

2^2i is the same as (e^2i)^ln2 which is (cos(2) + isin(2))^ln2

a simple way to convince you of it is to write X = 2^2i
then lnX = ln (2^2i) = 2iln2

so X = e^(2iln2) = (e^2i)^ln2

Answer 7

there's no such thing as a minus square root of a number as it is a number times itself (the same number) e.g - i ^2 = a positive and a positive x a positive also =s a positive

Answer 8

(i)^1/2=[cospi/2 +1sinpi/2]^1/2
=cos(pi/4+isinpi/4)or
= [1+i]/sqrt(2)
(i)^1/2= [cos pi/2-isinpi/2]^1/2
=[cos pi/4- isinpi/4].=[1-i] / sqr(2). ANS.

Answer 9

2^2i=4i

sq rt of i=.7071067812+.7071067812i

sq rt of -i = 7071067812-.7071067812i

and for everyone that said i is unknown..no its not, i is simply not a variable. it is the sq rt of -1, which is not possible, imaginary number, so they created a symbol (i)

Answer 10

2^2i = 4 * -1 = -4