Question

find the product of the complex number and its conjugate

(-1)-(squareroot 2i)

Answer 1

\((a+bi)(a-bi)=a^2+b^2\)
does this help?

Answer 2

kinda how do you use it?

Answer 3

u have \(-1-\sqrt{2}i\)
compare this with a+bi and tell what's a and b ?

Answer 4

((-1)-sqrt(2)i) *((-1)+sqrt(2)i)
foil
(-1)*(-1)+(1)*sqrt(2)i+(-1)sqrt(2)i-sqrt(2)^2*i*2

i^2 = -1

Answer 5

what @hartnn said. forget about i, forget about minus signs, etc. it is just \(a^2+b^2\) a positive real number

Answer 6

a= -1 and b=2?

Answer 7

try b again.

Answer 8

a is correct

Answer 9

square root 2?

Answer 10

\(b=-\sqrt{2}\) ok?

Answer 11

oh ok. so u can have a negative? is that the product we found?

Answer 12

the product is \(a^2+b^2\)
can u calculate it?

Answer 13

haha yes thanks!! you guys!

Answer 14

welcome :)
so what product u found?

Answer 15

1+ -2 which = -1

Answer 16

\((-\sqrt2)^2=2\)
not -2.

Answer 17

ait why? the negative doesnt get cancelled out

Answer 18

because (-1)^2=1
so \((-\sqrt2)^2=(-1)^2(\sqrt{2}^2)=2\)

Answer 19

ahhhh ok so the squared goes to both! thank you!!

Answer 20

yup,so whats final answer?

Answer 21

3

Answer 22

correct :)

Answer 23

thanks!!