Question

Find the product of the complex number and its conjugate.
1 + 3i

Answer 1

the conjugate of \(a+bi\) is \(a-bi\) and it is always true that
\[(a+bi)(a-bi)=a^2+b^2\]

Answer 2

forget about the \(i\) and forget about minus signs, in your case \(a=1,b=3\) and \(1^2+3^2=1+9=10\)

Answer 3

So is it 1+9i?

Answer 4

right..

Answer 5

That's the answer? Your ".." is misleading.

Answer 6

there is no "\(i\)" in the answer

Answer 7

no \(i\) and no \(-\) sign, only \(a^2+b^2\)

Answer 8

sorry my bad.... i^ 2 = 1

Answer 9

\((a+bi)(a-bi)=a^2+b^2\) so \((1+3i)(1-3i)=1^2+3^2=1+9=10\) a real number

Answer 10

1 + 9 = 10

Answer 11

1 + 9i
10
-8
1 - 9i
Are my answers, is it 10?

Answer 12

yes, it is 10

Answer 13

Thank you!