Question

Lily is practicing multiplying complex numbers using the complex number (2 + i).
To determine the value of left parenthesis 2 plus i right parenthesis squared , Lily performs the following operations:
i will post them after i post this.
Lily made an error. Explain Lily's error and correct the step which contains the error.

Answer 1

what do you think?

Answer 2

i think step 2 is wrong. is that right?

Answer 3

do you know how to multiply complex numbers?

Answer 4

do you know how to multiply binomials?

Answer 5

no i don't. I just started learning it.

Answer 6

do you know how to multiply binomials? like\[\left( x+1 \right)^2\]

Answer 7

do you distribute the 2 to the other numbers?

Answer 8

no... do you know the distributive property? like\[a \left( b+c \right)=a\cdot b+a\cdot c\]

Answer 9

yes

Answer 10

well, excellent! then you kinda already know how to multiply binomials (or any other nomials for that matter).

In order to multiply \(\left( x+ 1\right)^2\) we just use the distributive property:\[\left(x+1 \right)^2=\left(x+1 \right)\cdot \left(x+1 \right)\]This may seem confusing, but if we let \(a\) represent the \(\left( x+ 1\right)\) on the left then we get something that looks exactly like the distributive property:\[\left(x+1 \right)\cdot \left(x+1 \right)=a\cdot \left(x+1 \right)\]Can you perform this operation? If so, what do you get?

Answer 11

x2=a*x+1

Answer 12

not quite...\[a\cdot \left(x+1 \right)= a\cdot x + a\cdot 1\]We just used the distributive property.
But \(a\) isn't really \(a\), it's \(\left(x+1 \right)\). So look at what we get when we put it back in:\[a\cdot x + a\cdot 1 = \left(x+1 \right)\cdot x+\left(x+1 \right)\cdot 1\]Now, we just distribute again...\[\left(x+1 \right)\cdot x+\left(x+1 \right)\cdot 1=x\cdot x+x\cdot 1+x\cdot 1 + 1 \cdot 1 = x^2 +2x +1\]

Answer 13

okay i get it. but how does that work with my question that has an (i) in it

Answer 14

yes, same principle
\[\left( 2+i \right)^2=\left( 2+i \right)\cdot \left( 2+i \right)\]do the same thing and sue the distributive property. the only "new" thing here is that \(i^2=-1\).

Answer 15

oops... use not "sue"

Answer 16

okay i'll look at it and put in what i think it is.

Answer 17

okay, good luck!

Answer 18

(2+-1)*(2+-1)=4+(-1)*(-1)
its probably not right on how i distributed it.