Question

If we have a complex number as $z=1+2i$, then find the value of $\dfrac{1}{z}$.

Answer 1

Hint: Substitute the value of the given complex number in the given expression. 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 $\dfrac{1}{z}$. We observe that $z=1+2i$ is a complex number.
To do so, we will substitute $z=1+2i$ in the given expression.
 Thus, we have $\dfrac{1}{z}=\dfrac{1}{1+2i}$.
We know that we can simplify the expression of the form $\dfrac{1}{x+iy}$ by multiplying and dividing it by $x-iy$.
Substituting $x=1,y=2$ in the above expression, we can simplify it as $\dfrac{1}{1+2i}=\dfrac{1}{1+2i}\times \dfrac{1-2i}{1-2i}$.
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 $\dfrac{1}{1+2i}=\dfrac{1}{1+2i}\times \dfrac{1-2i}{1-2i}=\dfrac{1-2i}{{{1}^{2}}-{{\left( 2i \right)}^{2}}}$.
Thus, we have $\dfrac{1}{1+2i}=\dfrac{1}{1+2i}\times \dfrac{1-2i}{1-2i}=\dfrac{1-2i}{{{1}^{2}}-{{\left( 2i \right)}^{2}}}=\dfrac{1-2i}{1-4{{i}^{2}}}$.
We know that $i=\sqrt{-1}$. Thus, we have ${{i}^{2}}=-1$.
So, we have $\dfrac{1}{1+2i}=\dfrac{1}{1+2i}\times \dfrac{1-2i}{1-2i}=\dfrac{1-2i}{{{1}^{2}}-{{\left( 2i \right)}^{2}}}=\dfrac{1-2i}{1-4{{i}^{2}}}=\dfrac{1-2i}{1-4\left( -1 \right)}=\dfrac{1-2i}{1+4}=\dfrac{1-2i}{5}=\dfrac{1}{5}-\dfrac{2i}{5}$
Hence, the value of $\dfrac{1}{z}$ when $z=1+2i$ is $\dfrac{1-2i}{5}=\dfrac{1}{5}-\dfrac{2i}{5}$.

Note: We must keep in mind that $i=\sqrt{-1}$ is the root of unity. Thus, we have ${{i}^{2}}={{\left( \sqrt{-1} \right)}^{2}}=-1$. 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.