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4-1.Complex numbers
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If $arg\, z < 0$ then $arg\, (-z)\, -arg(z)$ is equal to
A
$\pi $
B
$-\pi $
C
$-\frac {\pi }{2}$
D
$\frac {\pi }{2}$
Solution
$\because \arg (z)<0$
$\because \arg (z)-\arg (-z)=-\pi$
Hence, $\arg (-\mathrm{z})-\arg (\mathrm{z})=\pi$
Standard 11
Mathematics
Similar Questions
Let $z$ be complex number satisfying $|z|^3+2 z^2+4 z-8=0$, where $\bar{z}$ denotes the complex conjugate of $z$. Let the imaginary part of $z$ be nonzero.
Match each entry in List-$I$ to the correct entries in List-$II$.
| List-$I$ | List-$II$ |
| ($P$) $|z|^2$ is equal to | ($1$) $12$ |
| ($Q$) $|z-\bar{z}|^2$ is equal to | ($2$) $4$ |
| ($R$) $|z|^2+|z+\bar{z}|^2$ is equal to | ($3$) $8$ |
| ($S$) $|z+1|^2$ is equal to | ($4$) $10$ |
| ($5$) $7$ |
The correct option is:
Let $z_k=\cos \left(\frac{2 k \pi}{10}\right)+ i \sin \left(\frac{2 k \pi}{10}\right) ; k =1,2, \ldots 9$.
| List $I$ | List $II$ |
| $P.$ For each $z_k$ there exists a $z_j$ such that $z_k \cdot z_j=1$ | $1.$ True |
| $Q.$ There exists a $k \in\{1,2, \ldots ., 9\}$ such that $z_{1 .} . z=z_k$ has no solution $z$ in the set of complex numbers. | $2.$ False |
| $R.$ $\frac{\left|1-z_1\right|\left|1-z_2\right| \ldots . .\left|1-z_9\right|}{10}$ equals | $3.$ $1$ |
| $S.$ $1-\sum_{k=1}^9 \cos \left(\frac{2 k \pi}{10}\right)$ equals | $4.$ $2$ |
Codes: $ \quad P \quad Q \quad R \quad S$
