Question

Can someone explain these two problems to me...

Answer 1

Multiply and simplify answer.
(5/3 +i)^2


I got 25/9 +10/3i+(-1)


second problem-
Divide

6+i/3-i

Answer 2

\[(\frac{5}{3 +i})^2\] do you mean ?

Answer 3

No (5/3 +i)^2

Answer 4

\[(\frac{ 5 }{ 3 }+i)^{2}\]like this?

Answer 5

Yes

Answer 6

\[i^2=-1\\(\frac{5}{3} +i)^2=\\\frac{25}{9} +2(\frac{5}{3})(i) +i^2=\\\frac{25}{9}+\frac{10}{3}i -1\]

Answer 7

Yes I have that

Answer 8

\[\frac{6+i}{3-i}=\\\frac{6+i}{3-i}*\frac{3+i}{3+i}=\\=\frac{(6+i)(3+i)}{3^2-i^2}=\\\]can you go on ?

Answer 9

Wait so thats it for the first problem????

Answer 10

no , first is almost done
just=st 25/9-1=(25-9)/9=16/9

Answer 11

??

Answer 12

25/9+10/3 i -1=
25/9-1 +10/3 i=
16/9 +10/3 i

Answer 13

So 16/9 + 10/3i is the final answer?

Answer 14

yes ,for 1st

Answer 15

Okay

Answer 16

do the second ?

Answer 17

Yes..

I have-

6+1/3-i * 3+i?3+i = 10+6i+3i+i^2/ 9+3i-3i-i^2 = 9+9i/10

Answer 18

\[\frac{ (6+i)(3+i) }{(3-i)(3+i) }=\frac{18+3i+6i+i^2}{9-(-1)}=\\frac{18-1+9i}{10}\]

Answer 19

18-1 +9i
--------
10

Answer 20

Where did you get the -1 for the top?

Answer 21

So the final answer is

17+9i/10 ??

Answer 22

yes

Answer 23

Alright.. Thank you.