Question

Can someone help me with this problem:

Which is the product of -√2/3 + i and its conjugate?
A. 13/9
B. 11/9
C. 5/3
D. -1/3

I don't understand how to do it.

Answer 1

@jhonyy9 @sshayer

Answer 2

Negative?

Answer 3

\((a+ib)(a-ib) = a^2 + b^2\)

Answer 4

Would it be -1/3?

Answer 5

is your problem \[-\frac{ \sqrt{2} }{ 3 }+\iota ~or~\frac{ -\sqrt{2} }{ 3+\iota }\]

Answer 6

you'll get

\((-\frac{\sqrt{2}}{3})^2 + (1)^2\)

\(= \frac{{2}}{9} + 1\)

Answer 7

@sshayer it's \[-\frac{ \sqrt{2} }{ 3 } + i\]

Answer 8

from the options it seems Irish boy's work is excellent.

Answer 9

I still don't understand at all

Answer 10

conjugate of \[a+ \iota b ~is~a- \iota b\]
\[\left( a+ \iota b \right)\left( a- \iota b \right)=a^2-b^2\]

Answer 11

Okay, yeah that makes sense

Answer 12

\[conjugate~of ~\frac{ -\sqrt{2} }{ 3 }+\iota~is~\frac{ -\sqrt{2} }{ 3 }-\iota \]

Answer 13

Alright, so what Irishboy said, was that then I would square everything and get 2/9+1

Answer 14

But that's not an option

Answer 15

\[\left( \frac{ -\sqrt{2} }{ 3 }+\iota \right)\left( -\frac{ \sqrt{2} }{ 3 }-\iota \right)=\left( -\frac{ \sqrt{2} }{ 3 } \right)^2-\iota^2\]
\[=\frac{ 2 }{ 9 }+1=\frac{ 2+9 }{ 9 }=\frac{ 11 }{ 9 }\]

Answer 16

you got it.

Answer 17

\[\iota^2=-1\]
-(-1)=+1

Answer 18

That helped so much. Thanks!!

Answer 19

yw