Question

What is conjugate of 2-i/(1-2i)^2

Answer 1

(1- 2i)^2 = 1 + (2i)^2 - 2*1*(2i) = 1 - 4 -4i = -3-4i
  2-i    =   i - 2 
-3-4i       3+4i
i - 2    *  3- 4i      =  3i - 4i^2 - 6 + 8i    = 11i - 2 
3+4i      3 - 4i                9 - 16*i^2             25
conjugate of 11i - 2    is  11i + 2
                      25               25

Answer 2

Answer:

The conjugate of the expression is $\frac{2-i}{(1-2i)^2}=-\frac{2}{25}-\frac{11}{25}i$

Step-by-step explanation:

Given : Expression $\frac{2-i}{(1-2i)^2}$

To find : What is the conjugate of the expression ?

Solution :

First we solve the expression,

$\frac{2-i}{(1-2i)^2}$

$=\frac{2-i}{1-4-4i}$

$=\frac{2-i}{-3-4i}$

$=\frac{i-2}{3+4i}$

Rationalize,

$=\frac{i-2}{3+4i}\times \frac{3-4i}{3-4i}$

$=\frac{3i-4i^2-6+8i}{3^2-(4i)^2}$

$=\frac{-2+11i}{9+16}$

$=\frac{-2+11i}{25}$

$=\frac{-2}{25}+\frac{11}{25}i$

The expression became $\frac{2-i}{(1-2i)^2}=-\frac{2}{25}+\frac{11}{25}i$

The conjugate of the complex number $a+ib$ is $a-ib$

So, The conjugate of the expression is $\frac{2-i}{(1-2i)^2}=-\frac{2}{25}-\frac{11}{25}i$