Notice 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
Perform the indicated operations and write the answer in the form a + bi, where a and b are real numbers.
Problem #1 (3+2i)^2
Problem #2 i^3 - i^2
9 answers
thanks...can I ask u to tell me if I'm correct when I do the problems please and repost back?
Reiny...
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
??????????????????????????
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
??????????????????????????
No, not right
(3+2i)^2= 9+12i+4i^2= 9+12i-4=5+12i
check that.
(3+2i)^2= 9+12i+4i^2= 9+12i-4=5+12i
check that.
is this the answer bob purley??? thanks for your help!!
You need to check it and answer for yourself it is it right.
it makes sense now...thanks!!
PROBLEM 2....
i^3 - i^2=
-i + the square root of -1^2=
-i + 1i
is this correct?
i^3 - i^2=
-i + the square root of -1^2=
-i + 1i
is this correct?
Perform all operations. Give your answer in
a + bi
form.
4(2 − i)
a + bi
form.
4(2 − i)