If \[\overrightarrow {\mathbf{r}} = x{\mathbf{\hat i}} + y{\mathbf{\hat j}},\] then
A. \[\overrightarrow \nabla \cdot \overrightarrow {\mathbf{r}} = 0{\text{ and }}\overrightarrow \nabla \left| {\overrightarrow {\mathbf{r}} } \right| = \overrightarrow {\mathbf{r}} \]
B. \[\overrightarrow \nabla \cdot \overrightarrow {\mathbf{r}} = 2{\text{ and }}\overrightarrow \nabla \left| {\overrightarrow {\mathbf{r}} } \right| = \overrightarrow {\mathbf{r}} \]
C. \[\overrightarrow \nabla \cdot \overrightarrow {\mathbf{r}} = 2{\text{ and }}\overrightarrow \nabla \left| {\overrightarrow {\mathbf{r}} } \right| = \frac{{{\mathbf{\hat r}}}}{r}\]
D. \[\overrightarrow \nabla \cdot \overrightarrow {\mathbf{r}} = 3{\text{ and }}\overrightarrow \nabla \left| {\overrightarrow {\mathbf{r}} } \right| = \frac{{{\mathbf{\hat r}}}}{r}\]
Answer: Option C
A. $$\frac{{1 + i}}{{\sqrt 2 }}a{\text{ and}} - \frac{{1 + i}}{{\sqrt 2 }}a$$
B. $$ia{\text{ and }} - ia$$
C. $$ia,\, - ia,\,\frac{{1 - i}}{{\sqrt 2 }}a{\text{ and}} - \frac{{1 - i}}{{\sqrt 2 }}a$$
D. $$\frac{{1 + i}}{{\sqrt 2 }}a,\, - \frac{{1 + i}}{{\sqrt 2 }}a,\,\frac{{1 - i}}{{\sqrt 2 }}a{\text{ and}} - \frac{{1 - i}}{{\sqrt 2 }}a$$
Which of the following functions of the complex variable z is not analytic everywhere?
A. ez
B. $$\sin \frac{{\text{z}}}{{\text{z}}}$$
C. e3
D. |z|3
A. \[\left( {1 - \sqrt 3 } \right){{{\bf{\hat i}}}^{\bf{'}}} + 3{{{\bf{\hat j}}}^{\bf{'}}} + \left( {1 + \sqrt 3 } \right){{{\bf{\hat k}}}^{\bf{'}}}\]
B. \[\left( {1 + \sqrt 3 } \right){{{\bf{\hat i}}}^{\bf{'}}} + 3{{{\bf{\hat j}}}^{\bf{'}}} + \left( {1 - \sqrt 3 } \right){{{\bf{\hat k}}}^{\bf{'}}}\]
C. \[\left( {1 - \sqrt 3 } \right){{{\bf{\hat i}}}^{\bf{'}}} + \left( {3 + \sqrt 3 } \right){{{\bf{\hat j}}}^{\bf{'}}} + 2{{{\bf{\hat k}}}^{\bf{'}}}\]
D. \[\left( {1 - \sqrt 3 } \right){{{\bf{\hat i}}}^{\bf{'}}} + \left( {3 - \sqrt 3 } \right){{{\bf{\hat j}}}^{\bf{'}}} + 2{{{\bf{\hat k}}}^{\bf{'}}}\]


Join The Discussion