Question

z= (1+7i)/(2-i)^2 then amp(z)=?

Answer 1

|2-i|^2=?

Answer 2

http://en.wikipedia.org/wiki/Complex_number#Polar_form

Answer 3

\[z= \frac{1+7i}{(2-i)^2} \]
start by expanding the square in the denominator
\[(2-i)^2=(2-i)\times(2-i)=\]

Answer 4

then

Answer 5

what did you get?

Answer 6

pls explain in details

Answer 7

\[(2−i)×(2−i)=2(2−i)-i(2−i)=\]

Answer 8

\[2(2-i)=2\times2+2\times-i=\]

Answer 9

\[i(2−i)=i\times2+i\times-i=\]

[ don't forget that i × i = i^2 = -1 ]

Answer 10

then

Answer 11

show me your working

Answer 12

i can't understand pls solve it in details

Answer 13

booo

Answer 14

if you just want solution ... use wolfram

http://www.wolframalpha.com/input/?i=%281%2B7i%29%2F%282-i%29%5E2

Answer 15

boooo

Answer 16

\[z\\= \frac{1+7i}{(2-i)^2}\\=\frac{1+7i}{(2−i)(2−i)}\\=\frac{1+7i}{2(2−i)-i(2−i)}
\\=\frac{1+7i}{2×2+2×−i-(i×2+i×−i)}\\=\frac{1+7i}{4-2i-(2i-i^2)}\\=\color{white}{\frac{1+7i}{4-2i-2i-1}}\\\color{white}{=\frac{1+7i}{3-4i}}\]

after you have simplified this denominator completely

use the complex conjugate to realize the denominator