Can you solve this? | iota maths problem | Oxford entrance exam question

[@camgere]

Or you could just start from -i = e^i(-pi/2)

[@andreaslarsson4076]

Well, just visualize it. Angles are added. 270/2 = 135 = 90+45. So that’s -1+i right there. Lengths are multiplied so scale by sqrt(2). (-1+i)/sqrt(2). The other solution is just a mirror. I.e. -45 degrees.

[@ianboard544]

If you think of -i in polar form it's a hell of a lot easier.

[@walterwen2975]

Oxford entrance exam question: √(- i) =?
- i = (- 2i)/2 = (1 - 2i - 1)/2 = (1 - 2i + i²)/2 = [(1 - i)²]/2 = [(1 - i)/√2]²
√(- i) = √{[(1 - i)/√2]²} = ± (1 - i)/√2 = ± (√2 - i√2)/2
Final answer:
√(- i) = (√2 - i√2)/2 or √(- i) = (- √2 + i√2)/2

[@RealQinnMalloryu4]

(x ➖ 1ix+1i).