If the equation \(a|z|^2+\overline{\bar{a} z+\alpha \bar{z}}+d=0\) represents a circle where \(a....

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The Equation zz* + az* + a*z + b = 0 is a Circle if |a|Β² is greater than equal to b. The Equation zz* + az* + a*z + b = 0 is a Circle if |a|Β² is greater than equal to b.
If the equation a|z|2 + Β―(¯𝜢 𝒛+πœΆΒ―π’›)+𝒅=𝟎 represents a circle where a, d are real constants, then If the equation a|z|2 + Β―(¯𝜢 𝒛+πœΆΒ―π’›)+𝒅=𝟎 represents a circle where a, d are real constants, then
if z=x+iy then show that if z=x+iy then show that 'z bar(z)+2(z+bar(z))+b=0 'where b in r represents a circle'
The equation `z(bar(z+1+isqrt3))+bar z (z+1+isqrt3) = 0` represents a circle with The equation `z(bar(z+1+isqrt3))+bar z (z+1+isqrt3) = 0` represents a circle with
Show z1,z2 inverse points w.r.t.circle zz bar+az bar+a bar z+r=0.iff z1z2 bar+a bar z1+az2 bar+r=0. Show z1,z2 inverse points w.r.t.circle zz bar+az bar+a bar z+r=0.iff z1z2 bar+a bar z1+az2 bar+r=0.
𝑧=𝑧 ̅ if and only if π‘§πœ–β„ 𝑧=𝑧 ̅ if and only if π‘§πœ–β„
50 Seconds: Why Real Part of z = (z + zΜ„)/2=Re(z) 50 Seconds: Why Real Part of z = (z + zΜ„)/2=Re(z)
Solving equations of the form k*|z - Ξ±| = |z - Ξ²|. Solving equations of the form k*|z - Ξ±| = |z - Ξ²|.