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Hint: Substitute the value of the given complex number in the given expression and simplify it using the algebraic identity ${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$. Further, simplify the expression of the form $\dfrac{1}{x+iy}$ by multiplying and dividing it by $x-iy$. Calculate the value of the expression using the algebraic identity $\left( x+y \right)\left( x-y \right)={{x}^{2}}-{{y}^{2}}$.

__Complete step-by-step solution -__

We know that $z=1+2i$. We have to calculate the value of $\dfrac{1}{{{z}^{2}}}$.

To do so, we will substitute $z=1+2i$ in the given expression and simplify it using the algebraic identity ${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$.

Thus, we have $\dfrac{1}{{{z}^{2}}}=\dfrac{1}{{{\left( 1+2i \right)}^{2}}}=\dfrac{1}{{{1}^{2}}+{{\left( 2i \right)}^{2}}+2\left( 1 \right)\left( 2i \right)}$.

We know that $i=\sqrt{-1}$. Thus, we have ${{i}^{2}}=-1$

So, we have \[\dfrac{1}{{{z}^{2}}}=\dfrac{1}{{{1}^{2}}+{{\left( 2i \right)}^{2}}+2\left( 1 \right)\left( 2i \right)}=\dfrac{1}{1+4{{i}^{2}}+4i}=\dfrac{1}{1-4+4i}=\dfrac{1}{-3+4i}\].

We know that we can simplify the expression of the form $\dfrac{1}{x+iy}$ by multiplying and dividing it by $x-iy$.

Substituting $x=-3,y=4$ in the above expression, we can simplify it as $\dfrac{1}{-3+4i}=\dfrac{1}{-3+4i}\times \dfrac{-3-4i}{-3-4i}$.

We know the algebraic identity $\left( x+y \right)\left( x-y \right)={{x}^{2}}-{{y}^{2}}$. So, we can simplify the above expression as $\dfrac{1}{-3+4i}=\dfrac{1}{-3+4i}\times \dfrac{-3-4i}{-3-4i}=\dfrac{-3-4i}{{{\left( -3 \right)}^{2}}-{{\left( 4i \right)}^{2}}}$.

Thus, we have $\dfrac{1}{-3+4i}=\dfrac{1}{-3+4i}\times \dfrac{-3-4i}{-3-4i}=\dfrac{-3-4i}{9-16{{i}^{2}}}=\dfrac{-3-4i}{9-16\left( -1 \right)}=\dfrac{-3-4i}{9+16}=\dfrac{-3-4i}{25}$.

Hence, the value of $\dfrac{1}{{{z}^{2}}}$ when $z=1+2i$ is $\dfrac{-3-4i}{25}$.

Note: We must keep in mind that $i=\sqrt{-1}$ is the root of unity. We can write any complex number in the form $a+ib$, where $ib$ is the imaginary part and $a$ is the real part. We can’t solve this question without using algebraic identities. We must simplify the complex part in the denominator of a fraction by rearranging the terms.

We know that $z=1+2i$. We have to calculate the value of $\dfrac{1}{{{z}^{2}}}$.

To do so, we will substitute $z=1+2i$ in the given expression and simplify it using the algebraic identity ${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$.

Thus, we have $\dfrac{1}{{{z}^{2}}}=\dfrac{1}{{{\left( 1+2i \right)}^{2}}}=\dfrac{1}{{{1}^{2}}+{{\left( 2i \right)}^{2}}+2\left( 1 \right)\left( 2i \right)}$.

We know that $i=\sqrt{-1}$. Thus, we have ${{i}^{2}}=-1$

So, we have \[\dfrac{1}{{{z}^{2}}}=\dfrac{1}{{{1}^{2}}+{{\left( 2i \right)}^{2}}+2\left( 1 \right)\left( 2i \right)}=\dfrac{1}{1+4{{i}^{2}}+4i}=\dfrac{1}{1-4+4i}=\dfrac{1}{-3+4i}\].

We know that we can simplify the expression of the form $\dfrac{1}{x+iy}$ by multiplying and dividing it by $x-iy$.

Substituting $x=-3,y=4$ in the above expression, we can simplify it as $\dfrac{1}{-3+4i}=\dfrac{1}{-3+4i}\times \dfrac{-3-4i}{-3-4i}$.

We know the algebraic identity $\left( x+y \right)\left( x-y \right)={{x}^{2}}-{{y}^{2}}$. So, we can simplify the above expression as $\dfrac{1}{-3+4i}=\dfrac{1}{-3+4i}\times \dfrac{-3-4i}{-3-4i}=\dfrac{-3-4i}{{{\left( -3 \right)}^{2}}-{{\left( 4i \right)}^{2}}}$.

Thus, we have $\dfrac{1}{-3+4i}=\dfrac{1}{-3+4i}\times \dfrac{-3-4i}{-3-4i}=\dfrac{-3-4i}{9-16{{i}^{2}}}=\dfrac{-3-4i}{9-16\left( -1 \right)}=\dfrac{-3-4i}{9+16}=\dfrac{-3-4i}{25}$.

Hence, the value of $\dfrac{1}{{{z}^{2}}}$ when $z=1+2i$ is $\dfrac{-3-4i}{25}$.

Note: We must keep in mind that $i=\sqrt{-1}$ is the root of unity. We can write any complex number in the form $a+ib$, where $ib$ is the imaginary part and $a$ is the real part. We can’t solve this question without using algebraic identities. We must simplify the complex part in the denominator of a fraction by rearranging the terms.

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