Question

How do you simplify $\left[ \dfrac{{{\left( 7+8i \right)}^{10}}}{{{\left( -8+7i \right)}^{11}}} \right]$ ?

Answer 1

Hint: In the given question we need to simplify the term $\left[ \dfrac{{{\left( 7+8i \right)}^{10}}}{{{\left( -8+7i \right)}^{11}}} \right]$which involves iota in the denominator and numerator. So, our main motive is to write it in the simplest form and to remove the iota from the denominator. Also, make sure that all complex terms are in one bracket and real terms in another one bracket.

Complete step-by-step solution:
According to the given question we need to simplify $\left[ \dfrac{{{\left( 7+8i \right)}^{10}}}{{{\left( -8+7i \right)}^{11}}} \right]$.
So, now we will replace minus sign with iota square at first as we know ${{i}^{2}}=-1$
Therefore, we get $\left[ \dfrac{{{\left( 7+8i \right)}^{10}}}{{{\left( {{i}^{2}}8+7i \right)}^{11}}} \right]$ and now taking out i common from denominator we get $\left[ \dfrac{{{\left( 7+8i \right)}^{10}}}{{{i}^{11}}{{\left( i8+7 \right)}^{11}}} \right]$.
Now, we can see that the terms inside the bracket are the same in denominator and numerator 7+8i.
Therefore, we get $\left[ \dfrac{1}{{{i}^{11}}\left( 7+8i \right)} \right]$
And now we know that we can simplify the denominator further by reducing the power by iota involved in denominator as $\left[ \dfrac{1}{-i\left( 7+8i \right)} \right]$and now multiplying with iota inside the bracket we get $\left[ \dfrac{1}{\left( -7i-8{{i}^{2}} \right)} \right]$and since we know that ${{i}^{2}}=-1$we get $\left[ \dfrac{1}{\left( -7i+8 \right)} \right]$.
Now, we are almost done what we need to do is we need to rationalise the denominator and we will get
$\begin{align}
  & \dfrac{1}{\left( -7i+8 \right)}\times \dfrac{\left( 8+7i \right)}{\left( 8+7i \right)} \\
 & \Rightarrow \dfrac{\left( 8+7i \right)}{113} \\
\end{align}$
Therefore, the simplified form of the given question is $\dfrac{\left( 8+7i \right)}{113}$.

Note: We need to remember in such questions that we need to remove minus signs using iota wisely. It is not just removing every minus sign and writing iota there which may lead to complex terms and we won’t be able to remove iota from the denominator and will move around the same steps of simplification back and forth.