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Or you could just start from -i = e^i(-pi/2)
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.
If you think of -i in polar form it's a hell of a lot easier.
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)/2Final answer: √(- i) = (√2 - i√2)/2 or √(- i) = (- √2 + i√2)/2
(x ➖ 1ix+1i).
Or you could just start from -i = e^i(-pi/2)
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.
If you think of -i in polar form it's a hell of a lot easier.
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
(x ➖ 1ix+1i).