If a complex number z is such that (7+i)(z+ba | vedclass

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(C) Let $z = x + iy$. Then $\bar{z} = x - iy$.$z + \bar{z} = 2x$ and $z - \bar{z} = 2iy$.Substituting these into the given equation:$(7 + i)(2x) - (4

Vedclass · Accepted Answer

$200$

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(C) Let $z = x + iy$. Then $\bar{z} = x - iy$.$z + \bar{z} = 2x$ and $z - \bar{z} = 2iy$.Substituting these into the given equation:$(7 + i)(2x) - (4 + i)(2iy) + 116i = 0$$(14x + 2ix) - (8iy - 2y) + 116i = 0$$(14x + 2y) + i(2x - 8y + 116) = 0$Equating real and imaginary parts to zero:$14x + 2y = 0 \Rightarrow y = -7x$$2x - 8y + 116 = 0$Substitute $y = -7x$ into the second equation:$2x - 8(-7x) + 116 = 0$$2x + 56x + 116 = 0$$58x = -116 \Rightarrow x = -2$Then $y = -7(-2) = 14$.Thus,$z\bar{z} = |z|^2 = x^2 + y^2 = (-2)^2 + 14^2 = 4 + 196 = 200$.

If a complex number $z$ is such that $(7+i)(z+\bar{z})-(4+i)(z-\bar{z})+116i=0$,then $z\bar{z}=$

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