Let α, β Be the Roots of the Equation x² − x + 2 = 0 With Im(α) > Im(β). Then α⁶ + α⁴ + β⁴ − 5α² Is Equal To
Given α, β are the roots of x²−x+2=0 with Im(α)>Im(β), find α⁶+α⁴+β⁴−5α² by repeatedly reducing powers using α²=α−2.
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Let z Be a Complex Number Such That |z + 2| = 1 and Im((z+1)/(z+2)) = 1/5. Then the Value of |Re(z + 2)| Is
Given |z+2|=1 and Im((z+1)/(z+2))=1/5, find |Re(z+2)| by extracting y directly from the imaginary-part condition and using the circle equation for x.
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If z Is a Complex Number Such That |z| ≥ 2, Then the Minimum Value of |z + 1/2| Is
Given |z|≥2, find the minimum value of |z+1/2| by writing z in polar form and minimising the distance from z to the point −1/2.
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Let z Be a Complex Number Such That the Real Part of (z − 2i)/(z + 2i) Is Zero. Then the Maximum Value of |z − (6 + 8i)| Is Equal To
Given the real part of (z−2i)/(z+2i) is zero, show z traces a circle of radius 2 about the origin, then find the maximum distance from that circle to (6,8).
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The Number of Complex Numbers z Satisfying |z| = 1 and |z/z̄ + z̄/z| = 1 Is
Find how many complex numbers z satisfy both |z|=1 and |z/z̄+z̄/z|=1 simultaneously, by reducing the condition to x²−y²=±1/2 on the unit circle.
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If S = {z ∈ C : |z − i| = |z + i| = |z − 1|}, Then n(S) Is
Find the number of complex numbers z satisfying |z−i|=|z+i|=|z−1| simultaneously, by recognising z as the circumcenter of the triangle formed by i, −i, and 1.
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If the Centre and Radius of the Circle |(z−2)/(z−3)| = 2 Are Respectively (α, β) and γ, Then 3(α + β + γ) Is Equal To
The locus |(z−2)/(z−3)|=2 in the complex plane is a circle with center (α,β) and radius γ. Find 3(α+β+γ) by converting to Cartesian form.
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If z₁, z₂ Are Two Distinct Complex Numbers Such That |z₁−2z₂|/|1/2−z₁z̄₂| = 2, Then
If z1, z2 are distinct complex numbers with |z1−2z2|/|1/2−z1z̄2|=2, determine which circle condition z1 and z2 must satisfy, by expanding both sides.
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Roots of 2z²−3z−2i=0 — Find 16·Re(E)·Im(E) for a Power-Sum Ratio
If α, β are roots of 2z²−3z−2i=0, find 16·Re(E)·Im(E) where E=(α^19+β^19+α^11+β^11)/(α^15+β^15), using a Newton’s-identity shortcut.
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Let α and β Be Two Roots of the Equation x² + 2x + 2 = 0. Then α¹⁵ + β¹⁵ Is Equal To
If α, β are the roots of x²+2x+2=0, find α¹⁵+β¹⁵ by converting the complex roots to polar form and using De Moivre’s theorem.