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Question: If we have a complex number as \(z=1+2i\), find the value of \(\dfrac{1}{{{z}^{2}}}\)....

If we have a complex number as z=1+2iz=1+2i, find the value of 1z2\dfrac{1}{{{z}^{2}}}.

Explanation

Solution

Hint: Substitute the value of the given complex number in the given expression and simplify it using the algebraic identity (a+b)2=a2+b2+2ab{{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab. Further, simplify the expression of the form 1x+iy\dfrac{1}{x+iy} by multiplying and dividing it by x−iyx-iy. Calculate the value of the expression using the algebraic identity (x+y)(x−y)=x2−y2\left( x+y \right)\left( x-y \right)={{x}^{2}}-{{y}^{2}}.

Complete step-by-step solution -
We know that z=1+2iz=1+2i. We have to calculate the value of 1z2\dfrac{1}{{{z}^{2}}}.
To do so, we will substitute z=1+2iz=1+2i in the given expression and simplify it using the algebraic identity (a+b)2=a2+b2+2ab{{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab.
Thus, we have 1z2=1(1+2i)2=112+(2i)2+2(1)(2i)\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=−1i=\sqrt{-1}. Thus, we have i2=−1{{i}^{2}}=-1
So, we have 1z2=112+(2i)2+2(1)(2i)=11+4i2+4i=11−4+4i=1−3+4i\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 1x+iy\dfrac{1}{x+iy} by multiplying and dividing it by x−iyx-iy.
Substituting x=−3,y=4x=-3,y=4 in the above expression, we can simplify it as 1−3+4i=1−3+4i×−3−4i−3−4i\dfrac{1}{-3+4i}=\dfrac{1}{-3+4i}\times \dfrac{-3-4i}{-3-4i}.
We know the algebraic identity (x+y)(x−y)=x2−y2\left( x+y \right)\left( x-y \right)={{x}^{2}}-{{y}^{2}}. So, we can simplify the above expression as 1−3+4i=1−3+4i×−3−4i−3−4i=−3−4i(−3)2−(4i)2\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 1−3+4i=1−3+4i×−3−4i−3−4i=−3−4i9−16i2=−3−4i9−16(−1)=−3−4i9+16=−3−4i25\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 1z2\dfrac{1}{{{z}^{2}}} when z=1+2iz=1+2i is −3−4i25\dfrac{-3-4i}{25}.

Note: We must keep in mind that i=−1i=\sqrt{-1} is the root of unity. We can write any complex number in the form a+iba+ib, where ibib is the imaginary part and aa 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.