Question

Can a complex number be written Cartesian form in terms of i to a power other than 1? And if so...

I came across the question:
`z_1 = -2 -2sqrt(3i)`
`z_2 = 3sqrt(3) + 3i`
Find `z_3 = z_1 z_2` in Cartesian form.

But I keep ending up with multiple coefficients in terms of i, but each to a different power, such as
`-6sqrt(3) + 6isqrt(3i) -18sqrt(i) -6i`.
How can this be written in Cartesian form properly, or did I do something wrong to reach this?

Answer 1

The square roots of `"i"` are `pm1/sqrt(2) (1+i)`. So `sqrt("i") = 1/sqrt(2)(1+"i")`.

Explanation:

Yes, this number can be written in Cartesian form. The problem at the moment is that you have a `sqrt("i")`.

Your problem seems to be with taking the square root of a complex number. There are two ways to do this: one with just knowledge of basic algebra and one (far more useful) with knowledge of Euler's identity.

Way 1: Basic algebra.

Let `z=sqrt("i")` and `z=a+b"i"`, where `a, b in RR`.

Then,

`(a+b"i")^2=i`,
`a^2+2ab"i"-b^2="i"`,
`a^2-b^2+(2ab)"i"=0+"i"`.

Equating coeffecients `a^2-b^2=0` and `2ab=1`.

`a^2-b^2=0 => a=pmb`.

Then `2a^2=pm1`. As `a in RR`, `a=b=1/sqrt(2)`. We conclude that the square roots of `"i"` are `pm1/sqrt(2) (1+i)`.

We can check this by squaring `1/sqrt(2) (1+"i")`.

Way 2: Eulers identity

Let `z` be the complex number you want to raise to an exponent. A complex number can be written as `z=re^("i"theta)`.

`re^("i"theta)=r(cos(theta)+"i"sin(theta))` by eulers identity.

The sine and cosine functions have periodicity `2pi`. So any complex number `z` can be written as `z=re^("i"theta + 2kpi)` with `k in ZZ`.

Then,

`z^n=r^(n)e^(n("i"(theta+2kpi)))` for `n in RR`.

So, for this specific example, let `z=i`. Then `abs(z)=r=sqrt(1^2)=1`.
The number `"i"` makes a anticlockwise angle of `pi/2` from the positive `x` axis.

Then we conclude `z="i"=e^("i"pi(1/2+2k))`.

`z^(1/2)=e^("i"pi/2(1/2+2k))`.

Then choose `k` such that we receive all the arguments between `-pi` and `pi`.

`k=0 => z^(1/2)=e^("i"pi/4)`,
`k=-1 => z^(1/2)=e^(-"i"(3pi)/4)`.

When we raise to a power `1/n` `n in ZZ` we expect `n` solutions for `z`. In this case `n=2` and we have found the two solutions. By writing `e^("i"theta)=cos(theta)+"i"sin(theta)` you will see that these two solutions for `z^(1/2)` give the square root of `"i"` as `pm 1/sqrt(2) (1+"i")`.

I hope you can see how this method is more applicable to finding the roots of complex numbers more generally (and demonstrates that they can always be written in Cartesian form).

I trust you can use `sqrt(3"i")=sqrt("3")sqrt("i")=sqrt("3")/sqrt("2")(1+"i")` to finish the question you asked!