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April 25, 2014

Posted by **Dawn** on Tuesday, February 9, 2010 at 9:56am.

Problem #1 (3+2i)^2

Problem #2 i^3 - i^2

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**Reiny**, Tuesday, February 9, 2010 at 10:45amNotice the following pattern of powers of i

i = √-1

i^2 = √-1√-1 = -1

i^3 =i(i^2) = i(-1) = -i

i^4 = (i^2)(i^2) = (-1)(-1) = +1

i^5 = i(i^4) = i(1) = i

Can you see the pattern

for your first question , just expand it like you would (a+b)^2, then replace the powers of i

for the second, you should be able to do it following the above patterns

- math -
**Dawn**, Tuesday, February 9, 2010 at 11:17amthanks...can I ask u to tell me if I'm correct when I do the problems please and repost back?

- math -
**Dawn**, Tuesday, February 9, 2010 at 11:32amReiny...

are these the correct answers??

problem #1: (3+2i)^2=

(a+b)^2=

(3+2times-1)=

(3+1)^2=

4^2= 16

problem #2: i^3-i^2=

a-b=

-i- -1

??????????????????????????

- math -
**bobpursley**, Tuesday, February 9, 2010 at 11:48amNo, not right

(3+2i)^2= 9+12i+4i^2= 9+12i-4=5+12i

check that.

- math -
**Dawn**, Tuesday, February 9, 2010 at 12:08pmis this the answer bob purley??? thanks for your help!!

- math -
**bobpursley**, Tuesday, February 9, 2010 at 12:44pmYou need to check it and answer for yourself it is it right.

- math -
**Dawn**, Tuesday, February 9, 2010 at 1:10pmit makes sense now...thanks!!

- math -
**Dawn**, Tuesday, February 9, 2010 at 2:25pmPROBLEM 2....

i^3 - i^2=

-i + the square root of -1^2=

-i + 1i

is this correct?

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