Let z - i be any complex number such that z-i z+i is a purely imaginary number then z+ 1…
Mathematics · JEE Main · NTA Exams — Complex Numbers and Quadratic Equations
Let z \(\neq\)- i be any complex number such that \(\frac{z-i}{z+i}\) is a purely imaginary number then \(z+\frac{1}{z}\) is:
- 0
- any non-zero real number other than 1.
- any non-zero real number.
- a purely imaginary number.
Answer
(C) any non-zero real number.
Sign up free on Edukali to view the step-by-step worked solution and practice thousands of similar questions.
Related practice questions
- If n is a positive integer, then (1+i)^ n +(1-i)^ -n is equal to :
- The number of complex numbers z such that | z – 1| = | z + 1 | = | z – i | equals
- The value of
- If ω (≠ 1) is a complex cube root of unity, the least value of n ∈ N for which (1+ ^ 2 )^ 7 = (1+ ^ 4 )^ 7 is
- If a, b ∈ R & the quadratic equation ax 2 – bx + 1 = 0 has imaginary roots then a + b + 1 is
- i^ 2 +i^ 4 +i^ 6 + upto (2 k+1) terms, k N is
- If P, Q, R, S are represented by the complex numbers 4 + i, 1 + 6i, - 4 + 3i, - 1 - 2i respectively, then…
- If z = x + iy and w = (1 - iz) / (z - i), then |w| = 1 implies that, in the complex plane
More Complex Numbers and Quadratic Equations questions · Browse all practice questions