Question

Explain in the form of $a+ib$ where $a,b\in R$ ,$i=\sqrt{-1}$ . State values of a and b.
(i) $\left( 1+2i \right)\left( -2+i \right)$
(ii) $\left( 1+i \right){{\left( 1-i \right)}^{-1}}$

Answer 1

Hint: In the given questions we are given some complex numbers which we need to simplify and write in such a way that the real and imaginary parts are written separately. Also, we are asked to write the real and imaginary parts.

Complete step-by-step solution:
According to the question, we are given some complex numbers which we need to simplify by rationalizing and representing in such a way that the real and imaginary parts are written separately.
Now, in part (i) we are given two complex factors and we need to find the product initially and then write the real part and imaginary part.
Now, $\left( 1+2i \right)\left( -2+i \right)=-2-4i+i+2{{i}^{2}}$
Now, we know that the value of ${{i}^{2}}$ is -1. Therefore, replacing the value in our gained expression we get $\left( 1+2i \right)\left( -2+i \right)=-2-4i+i-2$.
Now, adding the real parts and also the imaginary part we will get the simplified form as $-4-3i$ .
Therefore, $a=-4$ and $b=-3$ .
Now, similarly simplifying the (ii) part we get, $\left( 1+i \right){{\left( 1-i \right)}^{-1}}=\dfrac{1+i}{1-i}$
Now we rationalise the complex number by multiplying and dividing by the conjugate of $1-i$, hence we get $\dfrac{1+i}{1-i} \times \dfrac{1+i}{1+i} = \dfrac{\left({1+i}\right)^2}{1-i^2}$
Now, again substituting the value of ${{i}^{2}}$ we will get $2$ in denominator and in numerator it will be $1+i^2+2i$ now on simplification we get $i$ .
Therefore, in this part we have only the imaginary part and the real part is 0 in this.
Therefore, $a=0$ and $b=1$ .

Note: In such a type of question, mainly where fractional terms are also involved, we forget to rationalize them and then we majorly have to remove the iota from the denominator in order to keep the term simple.