Show that (1+i)^2 = 2i
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Step-by-step explanation:

As we all know, i is same as root of -1

let us expand what we have;

(1 + i)^2

= 1(1 + i) + i(1 + i)

= 1 + i + i + i^2

as i is square root of -1, the square of the square root of -1 is -1

Thus, we have

1 + 2i - 1

= 2i

we have three similar triangles, because each has a right angle and shares an angle.   let's write the angles in order: opposite to short leg, long leg, hypotenuse.

cab similar to bad similar to cbd

or as ratios,

ca: ab: cb = ba: ad: bd = cb: bd: cd

we also know

ac = ad + cd

(ad+cd): ab: cb = ba: ad: bd = cb: bd: cd

(ad+cd)/cb=cb/cd

we have cb=6, ad=5 and seek x=cd.

we reject the negative root and conclude x=4

answer: 4