Question

Simplify ((sqrt5) -2i)^2

Answer 1

It looks like this:
\[(\sqrt{5}-2i)^2\]

Answer 2

\[\left( a-b \right)^2=a^2+b^2-2ab\]

Answer 3

In general,

\(\color{black}{\displaystyle (a+bi)^2=a^2+2a(bi)+(bi)^2=a^2+2abi+(b^2)(i^2)=a^2+2abi-b^2}\).

Answer 4

I had it set up as (\[(\sqrt{5}-2i)(\sqrt{5}-2i)\]
and then I started multiplying it out and got stuck after \[\sqrt{25}\]

Answer 5

Give your best shot, and we will fix your work if anything ...

Answer 6

I'm not sure the rules for multiplying \[\sqrt{5} and -2i\]

Answer 7

\[\sqrt{5}*\sqrt{5}=\left( \sqrt{5} \right)^2=5\]
or \[\sqrt{25}=5\]

Answer 8

you are doing good.

Answer 9

\(\color{black}{\displaystyle (\color{red}{\displaystyle \sqrt{{\tiny~}5{\tiny~~}}}-\color{blue}{2i})^2=(\color{red}{\sqrt{{\tiny~}5{\tiny~~}}})^2-2(\color{red}{\sqrt{{\tiny~}5{\tiny~~}}})(\color{blue}{2i})+(\color{blue}{2i})^2}\)

Answer 10

sshayer mentioned this rule in his first reply

Answer 11

the first two terms shouldn't be hard, and as regards to the last term,
the last term is in a form \((bi)^2\), and this is how it is simplified:

\((bi)^2=b^2\times i^2=b^2\times (-1)\) (since \(i^2=-1\))
\(=-b^2\).

Answer 12

***
I'm not sure the rules for multiplying
\(\sqrt{5}\) and−2i
***
first, remember that -2i means -2 * i
so you multiply all three to get
\[ \sqrt{5} \cdot -2 \cdot i \]
usually, people write the -2 first, and leave out the multiply signs
\[ -2 \sqrt{5} i \]
but it means the same thing : -2* sqr(5) * i