# What the?

• January 22nd 2006, 11:22 AM
MOODYtarqi
What the?
Can anyone help me with these..?!

1) 3i/2+5i

and...

2) square root of -2(square root of -9 - square root of 7)

i would appreicate it if you can reply asap.

P.S: only reply if u KNOW the answer...thanks.
• January 22nd 2006, 11:49 AM
CaptainBlack
Quote:

Originally Posted by MOODYtarqi
Can anyone help me with these..?!

1) 3i/2+5i

This is ambiguous (you need to use brackets to make this unambiguous),
but I will presume you are asking for:

$\frac{3i}{2+5i}$

to be reduced to standard form $a+b.i$.

Multiply the top and bottom by the complex conjugate of the bottom:

$\frac{3i}{2+5i} = \frac{3i (2-5i)}{(2+5i)(2-5i)}$,

so:

$\frac{3i}{2+5i} = \frac{15+6i}{29}=\frac{15}{29}+\frac{6i}{29}$.

RonL
• January 22nd 2006, 11:51 AM
MOODYtarqi
thanks man you're smart . . . or I'm an idiot.

appreciate it.
• January 22nd 2006, 12:02 PM
CaptainBlack
Quote:

Originally Posted by MOODYtarqi

2) square root of -2(square root of -9 - square root of 7)

This again is ambiguous, it could mean either:

$\sqrt{-2(\sqrt{-9}-\sqrt{7}})$,

or:

$\sqrt{-2}(\sqrt{-9}-\sqrt{7})$.

I will assume that you mean the latter of these, then:

$\sqrt{-2}(\sqrt{-9}-\sqrt{7})=\sqrt{2}i(3i-\sqrt{7})$

so:

$\sqrt{-2}(\sqrt{-9}-\sqrt{7})=-3\sqrt{2}-\sqrt{14}i$

RonL
• January 22nd 2006, 12:04 PM
MOODYtarqi
thanks again. but what if the tell me "write it in standard form?"
• January 22nd 2006, 12:10 PM
CaptainBlack
Quote:

Originally Posted by MOODYtarqi
thanks again. but what if the tell me "write it in standard form?"

They have both been reduced to standard form (or near enough to get marked as correct).

RonL
• January 22nd 2006, 12:12 PM
MOODYtarqi
thanks man... have a Gday.