If x - iy x - y - r - y -1 - 2- the number of values of z satisfying - z - n - z 2- z - z - n -2-1 n - N - n -1 isA- 0B- 1C- 2D- 3

The given equation is | z |^n = (z^2 + z) | z |^n 2+ 1 ⇒ z^2 + z is real ⇒ z^2 + z = z̅^2 + z̅ ⇒ (z z̅)(z + z̅ + 1) = 0 ⇒ z = z̅ = x as z + z̅ + 1 ≠ 0(x ≠ 1 ...

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Byju's Answer

Standard XII

Mathematics

Properties of Conjugate of a Complex Number

If x + iy x ,...

Question

If $x + i y (x, y \in r, x \neq 1 / 2)$ , the number of values of z satisfying ${| z |}^{n} = (z^{2} + z) {| z |}^{n - 2} + 1 (n \in N, n > 1)$ is

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Solution

The correct option is A 1
The given equation is
${| z |}^{n} = (z^{2} + z) {| z |}^{n - 2} + 1$
$\Rightarrow z^{2} + z$ is real
$\Rightarrow z^{2} + z = {¯ z}^{2} + ¯ z$
$\Rightarrow (z - ¯ z) (z + ¯ z + 1) = 0$
$\Rightarrow z = ¯ z = x a s z + ¯ z + 1 \neq 0 (x \neq - 1 / 2)$
Hence, the given equation reduces to
$x^{n} = x^{n} + x {| x |}^{n - 2} + 1$
$\Rightarrow x {| x |}^{n - 2} = - 1$
$\Rightarrow x = - 1$
So number of solutions is 1.

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If x+iy(x,y∈r, x≠1/2), the number of values of z satisfying |z|n=(z2+z)|z|n−2+1(n∈N,n>1) is

The correct option is A 1The given equation is |z|n=(z2+z)|z|n−2+1 ⇒z2+z is real ⇒z2+z=¯z2+¯z ⇒(z−¯z)(z+¯z+1)=0 ⇒z=¯z=xasz+¯z+1≠0(x≠−1/2) Hence, the given equation reduces to xn=xn+x|x|n−2+1 ⇒x|x|n−2=−1 ⇒x=−1 So number of solutions is 1.

If $x + i y (x, y \in r, x \neq 1 / 2)$ , the number of values of z satisfying ${| z |}^{n} = (z^{2} + z) {| z |}^{n - 2} + 1 (n \in N, n > 1)$ is