Let z be a purely imaginary number such that {mathop{rm Im}nolimits} (z)

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4-1.Complex numbers

normal

Let $z$ be a purely imaginary number such that ${\mathop{\rm Im}\nolimits} (z) < 0$. Then $arg\,(z)$ is equal to

$\pi $

$\frac{\pi }{2}$

$0$

$ - \frac{\pi }{2}$

Solution

(d)Let $z = 0 + ib$, where $b < 0$. Then $z$ is represented by a point on $OY'$ (negative direction of $y – $axis), therefore $arg(z) = – \frac{\pi }{2}$.

Standard 11

Mathematics

The argument of the complex number $\sin \,\frac{{6\pi }}{5}\, + \,i\,\left( {1\, + \,\cos \,\frac{{6\pi }}{5}} \right)$ is

normal

Let $z$ be a complex number (not lying on $X$-axis) of maximum modulus such that $\left| {z + \frac{1}{z}} \right| = 1$. Then

hard

If $\frac{{z – \alpha }}{{z + \alpha }}\left( {\alpha \in R} \right)$ is a purely imaginary number and $\left| z \right| = 2$, then a value of $\alpha $ is

hard

(JEE MAIN-2019)

Which of the following are correct for any two complex numbers ${z_1}$ and ${z_2}$

normal

If $z_1, z_2, z_3$ $\in$ $C$ such that $|z_1| = |z_2| = |z_3| = 2$, then greatest value of expression $|z_1 – z_2|.|z_2 – z_3| + |z_3 – z_1|.|z_1 – z_2| + |z_2 – z_3||z_3 – z_1|$ is

normal

Let $z$ be a purely imaginary number such that ${\mathop{\rm Im}\nolimits} (z) < 0$. Then $arg\,(z)$ is equal to

Solution

Similar Questions

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