Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I have one problem with Complex numbers.

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

This is ok?

share|improve this question
    
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
add comment

2 Answers 2

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

share|improve this answer
    
$$(-6i)^2 = 36i$$ or -6 * (-1) = 6 ? –  lala23 Feb 14 '12 at 16:00
2  
@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
add comment

What we have on the left hand side is: $(-6i)^2 = (-6)^2(i)^2 = 36*-1 = -36.$

On right right side, we have $(1-6i)^2 = (1-6i)(1-6i) =$ ... $= -35 - 12i.$

Can you fill in the missing steps? Do you see why the two sides cannot be equal?

share|improve this answer
    
Yes, I understand. Thank you –  lala23 Feb 14 '12 at 16:40
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.