Question #252559

Given two complex numbers zz and ww such that z=(1+i)w+(3i)wz=(1+i)w+(3-i)\overline{w} , where wˉ\bar{w} is the conjugate of ww . Deduce what is w\overline{w} in terms of zz and z\overline{z}


1
Expert's answer
2021-10-21T14:31:07-0400

z=(1+I)w+(3-i) conjugate (w)

z=(1+i)(3i)wˉ...................(1)z=(1+i)(3-i)\bar{w}...................(1)


zˉ=(1+i)w+(3i)wˉ\\\bar{z}=\overline{(1+i)w+(3-i)\bar{w}}\\


=(1+i).w+(3i).wˉ=\overline{(1+i)}.\overline{w}+\overline{(3-i)}.\overline{\bar{w}}\\


=(1i).w+(3+i)w........................(2)=(1-i).\overline{w}+(3+i)w........................(2)


Multiplying (3+i) and (1+i) in equation (1) and (2), we get;

z=(1+i)w+(3i)wˉ×(3+i)zˉ=(1i)wˉ+(3+i)w×(1+i)z=(1+i)w+(3-i)\bar{w}\times(3+i)\\\bar{z}=(1-i)\bar{w}+(3+i)w\times(1+i)


(3+i)z=(1+i)(3+i)w+(3i)(3+i)wˉ............(iii)(1+i)zˉ=(1i)(1+i)wˉ+(3+i)(1+i)w............(iv)(3+i)z=(1+i)(3+i)w+(3-i)(3+i)\bar{w}............(iii)\\(1+i)\bar{z}=(1-i)(1+i)\bar{w}+(3+i)(1+i)w............(iv)


subtracting equation (4) from equation (3) we get


wˉ[(1+1)+(9+1)]=z(3+i)(1+i)zˉ\bar{w}[-(1+1)+(9+1)]=z(3+i)-(1+i)\bar{z}


wˉ=18[(3+i)z(1+i)zˉ\bar{w}=\frac{1}{8}[(3+i)z-(1+i)\bar{z}


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