If a 0 and z = frac{(1 + i)^2}{a - i},where | vedclass

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

Question

(B) Given $z = \frac{(1 + i)^2}{a - i}$.Since $(1 + i)^2 = 1 + 2i + i^2 = 1 + 2i - 1 = 2i$,we have $z = \frac{2i}{a - i}$.Multiply numerator and denom

Vedclass · Accepted Answer

$-\frac{1}{5} - \frac{3}{5} i$

Vedclass · Answer

(B) Given $z = \frac{(1 + i)^2}{a - i}$.Since $(1 + i)^2 = 1 + 2i + i^2 = 1 + 2i - 1 = 2i$,we have $z = \frac{2i}{a - i}$.Multiply numerator and denominator by the conjugate $(a + i)$:$z = \frac{2i(a + i)}{a^2 + 1} = \frac{2ai - 2}{a^2 + 1} = \frac{-2 + 2ai}{a^2 + 1}$.The magnitude is $|z| = \sqrt{\frac{(-2)^2 + (2a)^2}{(a^2 + 1)^2}} = \sqrt{\frac{4(1 + a^2)}{(a^2 + 1)^2}} = \sqrt{\frac{4}{a^2 + 1}} = \frac{2}{\sqrt{a^2 + 1}}$.Given $|z| = \sqrt{\frac{2}{5}}$,so $\frac{2}{\sqrt{a^2 + 1}} = \sqrt{\frac{2}{5}}$.Squaring both sides: $\frac{4}{a^2 + 1} = \frac{2}{5}$ $\Rightarrow 20 = 2(a^2 + 1)$ $\Rightarrow 10 = a^2 + 1$ $\Rightarrow a^2 = 9$.Since $a > 0$,$a = 3$.Substituting $a = 3$ into $z$: $z = \frac{-2 + 2(3)i}{3^2 + 1} = \frac{-2 + 6i}{10} = -\frac{1}{5} + \frac{3}{5}i$.Therefore,the conjugate $\overline{z} = -\frac{1}{5} - \frac{3}{5}i$.

If $a > 0$ and $z = \frac{(1 + i)^2}{a - i}$,where $i = \sqrt{-1}$,has magnitude $\sqrt{\frac{2}{5}}$,then $\overline{z}$ is equal to

Similar Questions

If $z$ is a complex number,then $z \cdot \overline{z} = 0$ if and only if

Let $z_1$ and $z_2$ be two complex numbers such that $z_1 + z_2$ and $z_1 z_2$ are both real. Then:

If $|z_1| = |z_2|$ and $\text{amp}(z_1) + \text{amp}(z_2) = 0$,then

If $z$ is a complex number,then which of the following is not true?

If $z$ is a complex number such that $|z| \geqslant 2$,then the minimum value of $|z + \frac{1}{2}|$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module