# What the?

• Jan 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.

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

appreciate it.
• Jan 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
• Jan 22nd 2006, 12:04 PM
MOODYtarqi
thanks again. but what if the tell me "write it in standard form?"
• Jan 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
• Jan 22nd 2006, 12:12 PM
MOODYtarqi
thanks man... have a Gday.