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Take the square root of 2i

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...

10 months ago

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