What is the square of 4-7i?

Question

zepdrix · Answer

Hey Rose c: Welcome to OpenStudy.
So when we square a binomial, we get the first term squared, twice product of both, and the second term squared.$$\Large\rm (4-7\mathcal i)^2\quad=\quad (4)^2+2(-7\mathcal i\cdot 4)+(-7\mathcal i)^2$$

zepdrix · Answer

So we need to simplify a bit from there.

zepdrix · Answer

What do you think Rosey?
You can handle the first term I'm sure,
how bout the other two?

anonymous · Answer

Yes but not the other two i need help with the whole problem

zepdrix · Answer

The middle term simplifies down a bit if we just multiply the 2, 7 and 4.$$\Large\rm (4-7\mathcal i)^2\quad=\quad 16-2\cdot7\cdot4\cdot\mathcal i+(-7\mathcal i)^2$$Which gives us something like uhhhh,$$\Large\rm (4-7\mathcal i)^2\quad=\quad 16-56\mathcal i+(-7\mathcal i)^2$$ya?
Then for the last term...

zepdrix · Answer

$$\Large\rm (-7\mathcal i)^2=(-7)^2\mathcal i^2$$Remember that you're square the negative on the 7 as part of the multiplication, so that should be come positive.
Remember what happens when you multiply i by itself?

anonymous · Answer

ok thanks