Let a = text{Im}left( frac{1 + z^2}{2iz} righ | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Let $z = x + iy$. Since $|z| = 1$,we have $x^2 + y^2 = 1$.Consider the expression $\frac{1 + z^2}{2iz} = \frac{1 + (x + iy)^2}{2i(x + iy)} = \frac

Vedclass · Accepted Answer

$(-1, 1)$

Vedclass · Answer

(A) Let $z = x + iy$. Since $|z| = 1$,we have $x^2 + y^2 = 1$.Consider the expression $\frac{1 + z^2}{2iz} = \frac{1 + (x + iy)^2}{2i(x + iy)} = \frac{1 + x^2 - y^2 + 2ixy}{2ix - 2y} = \frac{(1 + x^2 - y^2) + 2ixy}{-2y + 2ix}$.Since $x^2 + y^2 = 1$,we have $1 - y^2 = x^2$. Substituting this into the numerator:$\frac{(x^2 + x^2) + 2ixy}{-2y + 2ix} = \frac{2x^2 + 2ixy}{2i(x + iy)} = \frac{2x(x + iy)}{2i(x + iy)} = \frac{x}{i} = -ix$.Thus,$a = 	ext{Im}(-ix) = -x$.Since $|z| = 1$,$x$ ranges from $-1$ to $1$. Given $z 
e \pm 1$,$x 
e \pm 1$.Therefore,$x \in (-1, 1)$,which implies $a = -x \in (-1, 1)$.Hence,the set $A = (-1, 1)$.

Let $a = \text{Im}\left( \frac{1 + z^2}{2iz} \right)$,where $z$ is any non-zero complex number. The set $A = \{ a : |z| = 1 \text{ and } z \ne \pm 1 \}$ is equal to

Similar Questions

The locus of the point $z=x+iy$ satisfying the equation $\left|\frac{z-1}{z+1}\right|=1$ is given by :

If $z = x + iy$ and the point $P$ in the Argand diagram represents $z$,then the locus of the point $P$ satisfying the equation $2|z - 2 - 3i| = 3|z - 2 + i|$ is a circle with centre

The points in the Argand plane given by $Z_1 = -3 + 5i$,$Z_2 = -1 + 6i$,$Z_3 = -2 + 8i$,and $Z_4 = -4 + 7i$ form a:

If $z=x+iy$ satisfies the condition $|z+1|=1$,then $z$ lies on the

Let $z$ be a complex number. Then the angle between vectors $z$ and $-iz$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module