Question

Multiply
(5-3i)(6+4i)

Simplify
i^77

radical -60

Answer 1

The first one, multiply the binomial the way you are used to do it, and in the end substitute for

\[ i^2 = -1\]

Answer 2

Instead of solving this question lemme ask you one thing :

what is (x-1)(y+2) ?

Answer 3

second one is just playing with cyclic expressions

\[ i^2 = -1 \\i^3=-i
\\i^4=1\]

Answer 4

@kenya.porter .. ?

Answer 5

I wont (-: Fortunately there are no solutions mentioned yet, only ways.

Answer 6

:) @kenya.porter please reply ..

Answer 7

xy+2x-y-2

Answer 8

very good .. so you did like this :
\[\large{(x-1)(y+2)=x(y)+x(2)-1(y)-1(2)=xy+2x-y-2}\]
right?

Answer 9

yes

Answer 10

ok so apply the same process in the given question ..
\[\large{(5-3i)(6+4i)=?}\]

can u tell me what r u getting?

Answer 11

42+2i

Answer 12

\[\large{30+20i-18i-12i^2}\]
\[\large{30+2i+12}\]
\[\large{42+2i}\]

very good so this is your answer ..

Answer 13

so shall i move on to second one @kenya.porter ??

Answer 14

yes,thank you so much i would appreciate it.

Answer 15

\[\large{i^{77}}\]
\[\large{(i^7)^{11}}\]

\[\large{i^7=i^3*i^4}\]
\[\large{i^7=-i * 1}\]
\[\large{i^7=-i}\]
hence \[\large{(i^7)^{11}=(-i)^{11}}\]
\[\large{(-i)^{11}=(-1)*(i)^{11}=-1*i^3*i^4*i^4}\]
\[\large{-1*-i*1*1}\]
\[\large{i}\]

Answer 16

got it @kenya.porter ?

Answer 17

yes i do

Answer 18

any more problems? @kenya.porter

Answer 19

yes radical -60

Answer 20

ok !! :
\[\large{i^{-60}}\]
\[\large{\frac{1}{i^{60}}}\]
\[\large{\frac{1}{(i^5*i^{10})^4}}\]
\[\large{\frac{1}{(i*-i)^4}}\]
\[\large{\frac{1}{(-i)^4}}\]
\[\large{\frac{1}{1}}\]
\[\large{ 1}\]

Answer 21

Introduction to complex numbers : http://openstudy.com/study#/updates/4fe196d3e4b06e92b870a14a

Practicing on complex numbers : http://openstudy.com/study#/updates/4fe5a252e4b06e92b873c296

best of luck..

Answer 22

ok thank you