Take the square root of 2i

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As with much of complex number the trick here is to change forms to polar representation.

If you think of an argand diagram the number i will be represented as a point straight up on the imaginary axis a distance 2 from the origin.

It can therefore be represented as 2i = 2*e^(iπ/2)

From here it's easy! Just  apply the same indices rules that you have grown so familiar with. 

2 goes to the square root of 2, e^(iπ/2) goes to e^(iπ/4).

so we have the expression (2i)^(1/2) = (2)^(1/2)*(iπ/4)

And now convert back to standard form!

We know the magnitude is square root 2, and the arguement is π/4. Imagined on the argand diagram this is a line slanting at 45 degrees to the horizontal.

We can use the identity e^(iθ) =cos(θ) + i*sin(θ)

cos(π/4)=sin(π/4)= 2^(-1/2)

Thankfully the square roots of 2 cancel (Careful! they will not allways do this!) Therefore we reach the answer:

(2i)^(1/2) = 1 + i

which is satisfyingly elegant

Simon T. GCSE Further Mathematics  tutor, A Level Further Mathematics...

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