Question

what is the conjugate of (1+2i/2+i) hole square​

Answer 1

We have to find the conjugate of,

$\left(\dfrac{1+2\iota}{2+\iota}\right)^2$

The given no. is not in standard complex number form, i.e.,  $a+b\iota$

So we have to convert it into standard form first.

First we just take the square.

$\left(\dfrac{1+2\iota}{2+\iota}\right)^2\ =\ \dfrac{(1+2\iota)^2}{(2+\iota)^2}\ =\ \dfrac{1+4\iota-4}{4+4\iota-1}\ =\ \dfrac{-3+4\iota}{3+4\iota}$

And now we multiply both numerator and denominator by the conjugate of the denominator.

$\dfrac{-3+4\iota}{3+4\iota}\ =\ \dfrac{(-3+4\iota)(3-4\iota)}{(3+4\iota)(3-4\iota)}\ =\ \dfrac{-9+12\iota+12\iota+16}{9+16}\\ \\ \\ =\ \dfrac{7+24\iota}{25}\ =\ \dfrac{7}{25}+\dfrac{24}{25}\ \iota$

Now we got it in standard form.

So we can say that its conjugate is  $\mathbf{\dfrac{7}{25}-\dfrac{24}{25}\ \iota}$

$\begin{aligned}&\dfrac{7}{25}-\dfrac{24}{25}\ \iota\\ \\ \Longrightarrow\ \ &\dfrac{7-24\iota}{25}\\ \\ \Longrightarrow\ \ &\dfrac{-9-12\iota-12\iota+16}{9+16}\\ \\ \Longrightarrow\ \ &\dfrac{(-3-4\iota)(3+4\iota)}{(3-4\iota)(3+4\iota)}\\ \\ \Longrightarrow\ \ &\dfrac{-3-4\iota}{3-4\iota}\\ \\ \Longrightarrow\ \ &\dfrac{1-4\iota-4}{4-4\iota-1}\\ \\ \Longrightarrow\ \ &\dfrac{(1-2\iota)^2}{(2-\iota)^2}\\ \\ \Longrightarrow\ \ &\mathbf{\left(\dfrac{1-2\iota}{2-\iota}\right)^2}\end{aligned}$

$\text{Therefore,}\\ \begin{center}\large\boxed{\overline{\left(\dfrac{1+2\iota}{2+\iota}\right)^2}\ =\ \left(\dfrac{1-2\iota}{2-\iota}\right)^2}\end{center}$

We could directly change the sign, couldn't we?!