Question

For $z\ne 0$, define $\log z=\log |z|+i\left(argz\right)$

where $-\pi <arg\left(z\right)\le \pi$ i.e. $arg\left(z\right)$ stands for the principal argument of $z$.

Then $z\log \left(e^{x+iy}\right)$ equals :

• A. $e^x+arg(x+iy)$
• B. $z(x+iy)$
• C. $x+iy+2k\pi ,k\in I$
• D. None of these

Answer 1

The correct option is B $z(x+iy)$

$zlog(e^x.e^{iy})$

$=zlog\left(e^x\right)+zlog\left(e^{iy}\right)$

$=zx+z\left(iy\right)$

$=z(x+iy)$