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Eleven Standard >> Conjugate of a complex number and its properties

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Conjugate of a complex number:

Let z \(\in\) c and z=a+ib, then conjugate of z, denoted by \(\overline{z}\)=a-ib

Example: z=2-i, then \(\overline{z}\)=2+i
         z\(\overline{z}\)=(2-i)(2+i)
                          =\(2^{2}-(i)^{2}\)
                          =4+1=5 \(\in\) R
Conjugate of -3i+2 is 3i+2
Conjugate of -5-3i is -5+3i

Properties of conjugate:

If \(z_{1}\) and \(z_{1}\) complex number such that
\(z_{1}\)=x+iy and \(z_{2}\)=a+ib where i=\(\sqrt{-i}\), then

(i) \(\overline{(\overline{z_{1}})}\)=\(z_{1}\)

    Proof: \(\overline{(\overline{z_{1}})}\)=x+iy=\(z_{1}\)

(ii) \(\overline{z_{1}+z_{2}}\)=\(\overline{z_{1}}+\overline{z_{2}}\)

    Proof: \(z_{1}\)+\(z_{2}\)=x+iy+a+ib=(x+a)+i(y+b)
           Now, \(\overline{(\overline{z})}\)=(x+a)-i(y+b)
                =(x-iy)+(a-ib)=\(\overline{z_{1}}+\overline{z_{2}}\)
                

(iii) \(\overline{(z_{1}z_{2})}=\overline{z_{1}}.\overline{z_{2}}\)

    Proof: z_{1}z_{2}=(x+iy)(a+ib)=ax+ibx-by
               =(ax-yb)+i(bx+ay)

         Now, \(\overline{z_{1}}.\overline{z_{2}}\)=(x+iy)(a-ib)
                                                   =(ax-by)+i(ay+bx)

(iv) \(\overline{\Big(\frac{z_{1}}{z_{2}}\Big)}\)=\(\frac{\overline{z_{1}}}{\overline{z_2}}\) provide \(z_{2}\neq0\)
 
    Let, \(\frac{z_{1}}{z_{2}}\)=\({z_{3}}\)
 \(\Rightarrow\) \(z_{1}=z_{2}z_{3}\)
 \(\Rightarrow\) \( \overline{z_{1}}=\overline{z_{2}}\overline{z_{3}}\)
 \(\Rightarrow\) \(\frac{\overline{z_{1}}}{\overline{z_2}}\)=\(\overline{z_{3}}\)=\(\overline{\Big(\frac{z_{1}}{z_{2}}\Big)}\)

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