How do you multiply (2-3i)^2

Question

solomonzelman · Answer

Wait, you are doing complex and imaginary numbers right now, or
just algebra or pre-algebra?

anonymous · Answer

i is the imaginary number, its algebra 2.

solomonzelman · Answer

Okay, good...

solomonzelman · Answer

there is a rule:  $\large\color{black}{(a-b)^2=a^2-2ab+b^2 }$ $\LARGE\color{white}{ \rm   \left|   \right|  }$
you know that $\large\color{black}{i=\sqrt{-1} }$$\LARGE\color{white}{ \rm   \left|   \right|  }$
and just like $\large\color{black}{\sqrt{a} \times \sqrt{a}=a}$ so is,$\LARGE\color{white}{ \rm   \left|   \right|  }$ $\large\color{black}{i \times i}$  (which is $\large\color{black}{\sqrt{-1} \times \sqrt{-1}}$) is$\LARGE\color{white}{ \rm   \left|   \right|  }$
equal to $\large\color{black}{-1.}$$\LARGE\color{white}{ \rm   \left|   \right|  }$
$\large\color{black}{\sqrt{a} \times \sqrt{a}=a}$$\LARGE\color{white}{ \rm   \left|   \right|  }$

solomonzelman · Answer

the last line shouldn't be there, read everything without the very last equation, please.

solomonzelman · Answer

$\large\color{black}{(2-3i)^2=(2)^2-2(2)(3i)+(3i)^2}$

solomonzelman · Answer

if you have any questions after you are done reading, please ask.