Question

Value of complex number $\left[i^{23}+\left(\dfrac{1}{i}\right)^{29}\right]^2$ is equal to?

Answer 1

We are required to solve the complex expression involving $i$ and inverse of $i$. For this we need to be aware about various properties of complex numbers and moreover we also need to be aware about some general identities of algebra. We will first resolve the inverse term and after that we will use some properties of complex numbers to solve this expression.

Complete step-by-step solution:
We have the following expression:
$\left[i^{23}+\left(\dfrac{1}{i}\right)^{29}\right]^2$
Now, we know that $i^2=-1$, we use this property and we write ${{i}^{23}}=\left( {{i}^{22}} \right)\times i={{\left( {{i}^{2}} \right)}^{11}}\times i$. Doing this, we get:
$i^{23}=-1\times i=-i$.
Now, we move to the next expression, and we first solve $\dfrac{1}{i}$. Multiplying numerator and denominator by $i$, we get:
$\dfrac{1}{i}=\dfrac{1}{i}\times\dfrac{i}{i}=\dfrac{i}{i^2}=\dfrac{i}{-1}=-i$
So, we have established that:
$\dfrac{1}{i}=-i$
Now, we raise this obtained quantity to the power of 29.
${{\left( \dfrac{1}{i} \right)}^{29}}={{\left( -i \right)}^{29}}={{\left( -i \right)}^{28}}\times -i={{\left( {{\left( -i \right)}^{4}} \right)}^{7}}\times -i=1\times -i=-i$
We have used a simple result:
${{i}^{4}}=1$
and this happens because iota squared is -1 so after squaring on both sides we will get the result.
So far, we get this:
${{\left[ {{i}^{23}}+{{\left( \dfrac{1}{i} \right)}^{29}} \right]}^{2}}={{\left[ -i-i \right]}^{2}}={{\left( -2i \right)}^{2}}=4{{i}^{2}}=-4$
Hence, the final result obtained is -4.

Note: Note that $i^2=-1$. It is common to see the square term in the exponent and forget the negative sign which would lead to an invalid result. Moreover, you can solve this question more quickly if you are just aware about simple properties of the imaginary $i$. You can directly keep the terms in the bracket and start solving and obtain the result in no time.