Question

simplify the expression square root -16/(3-3i)+(1-2i)

Answer 1

\(\bf \cfrac{\sqrt{-16}}{(3-3i)+(1-2i)}?\)

Answer 2

CORRECT

Answer 3

\(\bf \cfrac{\sqrt{-16}}{(3-3i)+(1-2i)}\implies \cfrac{\sqrt{-1\cdot 4^2}}{3-3i+1-2i}
\\ \quad \\
\cfrac{\sqrt{-1}\cdot \sqrt{4^2}}{4-5i}\implies \cfrac{ 4i}{4-5i}\)

and surely you've done this ones already.... get the conjugate and multiply top and bottom by that to get rid of the "i" at the bottom

Answer 4

ok would it be 8-4i/15?

Answer 5

well.. not quite.... did you find the conjugate of the denominator?

Answer 6

is it close or am i way off

Answer 7

?

Answer 8

well... hmmm how did you get it though?

Answer 9

any ideas on the conjugate of the denominator?

Answer 10

nope. can you tech me how to find the conjugate of the denominaotr

Answer 11

hmmm did you follow the above? confused about it yet?

Answer 12

yeah i understand all of the above. but there has to be another step

Answer 13

http://www.mathsisfun.com/algebra/images/conjugate.gif <--- see what the conjugate is? what do you think it'd be for say the denominator in this case?

Answer 14

so the sign just flips?

Answer 15

so the comjugate would be 4+5i?

Answer 16

yeap

Answer 17

ok thank you i got it

Answer 18

thus \(\bf \cfrac{ 4i}{4-5i}\cdot \cfrac{4+5i}{4+5i}\implies \cfrac{ 4i({4+5i})}{(4-5i)({4+5i})}
\\ \quad \\
recall\implies {\color{brown}{ (a-b)(a+b) = a^2-b^2}}\qquad thus
\\ \quad \\
\cfrac{ 4i({4+5i})}{(4-5i)({4+5i})}\implies \cfrac{ 4i({4+5i})}{(4)^2-(5i)^2}\implies \cfrac{ 4i({4+5i})}{16-5^2i^2}\)

can you see it from there?

Answer 19

-20-16i/9

Answer 20

well.... close.. lemme finish it

Answer 21

ok

Answer 22

\(\bf \cfrac{ 4i}{4-5i}\cdot \cfrac{4+5i}{4+5i}\implies \cfrac{ 4i({4+5i})}{(4-5i)({4+5i})}
\\ \quad \\
recall\implies {\color{brown}{ (a-b)(a+b) = a^2-b^2}}\qquad thus
\\ \quad \\
\cfrac{ 4i({4+5i})}{(4-5i)({4+5i})}\implies \cfrac{ 4i({4+5i})}{(4)^2-(5i)^2}\implies \cfrac{ 4i({4+5i})}{16-5^2i^2}
\\ \quad \\
\cfrac{16i+20i^2}{16-25i^2}
\\ \quad \\
recall\implies {\color{brown}{ i^2\to -1}}\qquad thus
\\ \quad \\
\cfrac{16i+20i^2}{16-25i^2}\implies \cfrac{16i+20(-1)}{16-25(-1)}\implies \cfrac{-20+16i}{16+25}
\\ \quad \\
\cfrac{-20+16i}{41}\)