Question

Can sombody please help me simplify this expression.?
-5+i over 2i

Answer 1

\(\large \frac{-5+i}{2i} \)
\(\large \frac{-5+i}{2i}\cdot \color {red}{ \frac{2i}{2i}} \)

can you multiply the top first?

Answer 2

ummmm, is it -10+2i over 4i??

Answer 3

I'm sorry, I don't really know how to do this/:

Answer 4

\(\large \frac{-5+i}{2i}\cdot \color {red}{ \frac{2i}{2i}} \)
= \(\large \frac{\color {red}{2i}(-5+i)}{\color {red}{2i}\cdot 2i} \)
= \(\large \frac{\color {red}{2i}(-5)+\color {red}{2i}(i)}{4i^2} \)
can you finish?

Answer 5

-10i+2i over 4i^2 ?

Answer 6

@dpaInc where'd you go?

Answer 7

you should have -10i + 2i^2 over 4i^2....

now make a substitution using the fact that i^2 = -1...

Answer 8

so, the answer is -12i over -4......?

Answer 9

@dpaInc ?

Answer 10

by "simplify" you mean write in standard form as \(a+bi\)?

Answer 11

Idk/: The question just says "Simplify the expression"

Answer 12

multiply top and bottom by \(i\) and recall that \(i^2=-1\) that gives you
\[\frac{-5+i}{2i}\times \frac{i}{i}=\frac{-5i+i^2}{2i^2}=\frac{-5i-1}{-2}\]

Answer 13

now you can rewrite as
\[\frac{1}{2}-\frac{5}{2}i\] if you like

Answer 14

there is no such mathematical operation as "simplify"
it is the lazy math teacher way of saying "give me the answer i want"
there is however standard form of a complex number, and that is \(a+bi\) where \(a\) and \(b\) are real numbers

Answer 15

actually i made a mistake above , answer should be
\[\frac{1}{2}+\frac{5}{2}i\]

Answer 16

Ok thank you so much! That helped me understand it a whole lot better.

Answer 17

yw