Question

Consider the following.1. $z\overline{z}=|z{|}^2$2. $z^{-1}=\frac{z}{|z{|}^2}$, where $z=$ complex numberWhich of the above statement is/are correct?

• (A) Only 1
• (B) Only 2
• (C) Both 1 and 2
• (D) Neither 1 nor 2

Answer 1

The Correct Option is A

Explanation:
Let, Complex number, $z=a+ib$Conjugate of complex number $=\overline{z}=a-ib$So, $z\overline{z}=(a+ib)(a-ib)$$\Rightarrow z\overline{z}=a^2-(ib{)}^2$$\Rightarrow z\overline{z}=a^2-i^2(b{)}^2$$\Rightarrow z\overline{z}=a^2-(ib{)}^2$$\Rightarrow z\overline{z}=a^2+b^2$...(1) $(\because i^2=-1)$As we know,Modulus of complex number is given by, $|z|=\sqrt{(a^2+b^2)}$On squaring both sides, we get$|z{|}^2={\left(\sqrt{a^2+b^2}\right)}^2$$\therefore |z{|}^2=a^2+b^2$...(2)Comparing equation (1) and (2), we get$z\overline{z}=|z{|}^2$Now,$z^{-1}=\frac{1}{z}=\frac{1}{a+ib}$$\Rightarrow z^{-1}=\frac{1}{a+ib}\times \frac{a-ib}{a-ib}$$\Rightarrow z^{-1}=\frac{1}{a+ib}\times \frac{a-ib}{a-ib}$$\Rightarrow z^{-1}=\frac{a-ib}{a^2-(ib{)}^2}$$\Rightarrow z^{-1}=\frac{a-ib}{a^2-i^2{b}^2}$$\Rightarrow z^{-1}=\frac{a-ib}{a^2-(-1)b^2}[\because i^2=-1]$$\Rightarrow z^{-1}=\frac{a-ib}{a^2+b^2}$$\therefore z^{-1}=\frac{\overline{z}}{|z{|}^2}\ne \frac{z}{|z{|}^2}$So, only statement 1 is correct.Hence, the correct option is (A).