Squares of Complex numbers

I have one problem with Complex numbers.

$$(-6i)^2 = (1-6i)^2$$

This is ok?

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For the right-hand side, expand like you do with $(a+b)^2=a^2+2ab+b^2$, or better for us, $a^2+b^2+2ab$. Let $a=1$, $b=-6i$. Then $a^2+b^2=1^2+(-6i)^2= -35$, and $2ab=-12i$. So right-hand side is $-(35+12i)$. –  André Nicolas Feb 14 '12 at 16:06
What have you tried? –  lhf Feb 14 '12 at 16:06
More generally, $(-z)^2 = (1-z)^2$ iff $z=1/2$. –  lhf Feb 14 '12 at 16:08
It is ok, when you would calculate $(-6i)^2-(1-6i)^2 \mod (1-12i)=0$. –  draks ... Feb 14 '12 at 19:03

It is not. The left-hand side is a real number but the right-hand side is not.

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$$(-6i)^2 = 36i$$ or -6 * (-1) = 6 ? –  lala23 Feb 14 '12 at 16:00
@lala23, $(-6i)^2 = (-6)^2(i^2) = (36)\cdot(-1) = -36$. –  lhf Feb 14 '12 at 16:01
@lala: Again, the left one's real, but the other is not. –  Ｊ. Ｍ. Feb 14 '12 at 16:03
What we have on the left hand side is: $(-6i)^2 = (-6)^2(i)^2 = 36*-1 = -36.$
On right right side, we have $(1-6i)^2 = (1-6i)(1-6i) =$ ... $= -35 - 12i.$