Find roots of z² - 4z + 5 = 0:
Az = 2 ± i
Bz = 2 ± 2i
Cz = 5 ± i
Dz = 1 ± 2i
Answer & Solution
Correct answer: A. z = 2 ± i
By quadratic formula: z = (4 ± √(16 - 20))/2 = (4 ± √(-4))/2 = (4 ± 2i)/2 = 2 ± i.
Related questions
The number of distinct $n$th roots of a non-zero complex number is:Using De Moivre's theorem, $(\cos\theta + i in\theta)^4$ equals:The modulus of $z = 5 + 12i$ is:The conjugate of $z = 3 - 4i$ is:Find (3 + 4i)/(1 + 2i):If z1, z2, z3 are vertices of an equilateral triangle, then:Solve: z² + z + 1 = 0:For roots of unity ω^n = 1: which is the SMALLEST n such that 1 + ω + ω² + ... + ω^(n-1) =