Verified answer

I'm a little rusty on this, but I think the way into it is to remember that all powers of complex numbers are cyclic when they're normalised to the unit circle.

3 - √3i has modulus 2√3

so if you factor that out, you can express it as 2√3*(√3/2 - i/2)

1 − i has modulus √2

so can be similarly expressed as √2*(√2/2 - (√2/2)i)

Now you'll find that powers of the bracketed complex numbers are cyclic, and can be expressed in very simple terms. For example:

(√3/2 - i/2)^3 = -i

(√3/2 - i/2)^6 = -1

(√3/2 - i/2)^12 = 1

(√2/2 - (√2/2)i)^2 = -i

(√2/2 - (√2/2)i)^4 = -1

(√2/2 - (√2/2)i)^8 = 1

and so on.

So now you can express higher powers in terms of these. For example:

(3 - √3i)^72

= (2√3*(√3/2 - i/2))^72

= ((2√3))^72 * 1

= (2^72)(3^36)

So (3 - √3i)^71 = (2^72)(3^36)/(3 - √3i) ... I've verified this result.

Do a similar analysis for (1 − i)^53 and you'll get a fairly simple expression for

((3 - √3i)^71) / ((1 − i)^53)

Use complex conjugates to get rid of complex numbers in the denominator, expand, and collect, and it'll ultimately come down to a form a + bi.