Question

simplify in form of a +bi

(2+3i)^-2 simplify in form of a +bi

(2+3i)^-2 @Mathematics

Answer 1

problem is typed right

Answer 2

not typed*

Answer 3

\[(2+3i)^-2\]

Answer 4

that's root -2

Answer 5

my guess is:\[\sqrt{2}+\sqrt{3i}, \sqrt{2}-i \sqrt{3}\]

Answer 6

nope when u have negative it becomes a fraction

Answer 7

(2+3i)^-2 = 1/(2+3i)^2 = 1/(-5 + 6i)

Answer 8

ur square root answer would be right if the exponent was 1/2

Answer 9

nocipher is wrong too

Answer 10

ah I gotcha, Nocipher I want to understand how you got what you did. where did the -5+6 come from?

Answer 11

its 1/(-5+12i)

Answer 12

explain?

Answer 13

Ah, Curry is right. Did the math in my head wrong

Answer 14

u treat the problem like a binomial not like an expression without variables

Answer 15

therefore use box method, except the asner will be denominator

Answer 16

Man you guys make me feel soo dumb when it comes to math.

Answer 17

Exponentiation of complex numbers by real numbers works just like it does for real numbers. The only case when exponentiation might get complicated is when you raise a number to a complex power.

Answer 18

so \[=1/2^2 + 3i^2= 1/(2+3i)(2+3i)\]

Answer 19

You can't distribute powers like that. Be careful (1/2^2) + 3i^2 = 1/4 - 3, not 1/((2+3i)(2+3i))

Answer 20

4+12i+9i^2

Answer 21

i^2 = 1

Answer 22

so 4-9=-5

Answer 23

-5+12i

Answer 24

i^2 = -1, not 1

Answer 25

yeah what I got was 1/12+12i, I'm supposed to clear the denominator too so I'm still workin.

Answer 26

1/(12+12i)

Answer 27

Ah, but you needed it in the form a+bi... so the answer we got was 1/(-5+12i). You can use the complex conjugate of the denominator, -5-12i, to further reduce your answer. So multiply 1/(-5+12i) by (-5-12i)/(-5-12i) and get (-5-12i)/169. Then in a+bi form that is (-5/169)+(-12/169)i.

Answer 28

AHHH it is negative 5!

Answer 29

thank you all very much! I understand the process much better now.