Solveeit Logo

Question

Question: If \(z=1+2i\), then find the value of \({{z}^{2}}\)....

If z=1+2iz=1+2i, then find the value of z2{{z}^{2}}.

Explanation

Solution

Hint: To evaluate the value of the given algebraic expression, 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. Calculate the value of the expression using the fact that ii is a root of unity.

Complete step-by-step solution -
We know that z=1+2iz=1+2i. We have to calculate the value of z2{{z}^{2}}.
We observe that z=1+2iz=1+2i is a complex number while z2{{z}^{2}} is an algebraic expression. We will evaluate the value of this expression at z=1+2iz=1+2i.
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 z2=(1+2i)2{{z}^{2}}={{\left( 1+2i \right)}^{2}}.
Simplifying the above expression using the algebraic identity, we have z2=(1+2i)2=12+(2i)2+2(2i)(1){{z}^{2}}={{\left( 1+2i \right)}^{2}}={{1}^{2}}+{{\left( 2i \right)}^{2}}+2\left( 2i \right)\left( 1 \right)
So, we have z2=(1+2i)2=12+(2i)2+2(1)(2i)=1+4i2+4i{{z}^{2}}={{\left( 1+2i \right)}^{2}}={{1}^{2}}+{{\left( 2i \right)}^{2}}+2\left( 1 \right)\left( 2i \right)=1+4{{i}^{2}}+4i.
We know that i=1i=\sqrt{-1}. Thus, we have i2=1{{i}^{2}}=-1.
Simplifying the above expression, we have z2=(1+2i)2=12+(2i)2+2(1)(2i)=1+4i2+4i=1+4(1)+4i{{z}^{2}}={{\left( 1+2i \right)}^{2}}={{1}^{2}}+{{\left( 2i \right)}^{2}}+2\left( 1 \right)\left( 2i \right)=1+4{{i}^{2}}+4i=1+4\left( -1 \right)+4i.
We can further solve the above expression to write it as z2=(1+2i)2=12+(2i)2+2(1)(2i)=1+4i2+4i=1+4(1)+4i=14+4i=3+4i{{z}^{2}}={{\left( 1+2i \right)}^{2}}={{1}^{2}}+{{\left( 2i \right)}^{2}}+2\left( 1 \right)\left( 2i \right)=1+4{{i}^{2}}+4i=1+4\left( -1 \right)+4i=1-4+4i=-3+4i.
Hence, the value of z2{{z}^{2}} when z=1+2iz=1+2i is 3+4i-3+4i.

Note: We must keep in mind that i=1i=\sqrt{-1} is the root of unity. It is a solution to the equation x2+1=0{{x}^{2}}+1=0. Thus, we have i2=(1)2=1{{i}^{2}}={{\left( \sqrt{-1} \right)}^{2}}=-1. 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 the algebraic identity.