Question

If $z=\frac{1+7i}{3-4i}$, then arg $z=1$.

• A. True
• B. False

Answer 1

The correct option is B False

$z=\frac{1+7i}{3-4i}$

$=\frac{1+7i}{3-4i}\times \frac{3+4i}{3+4i}$

$=\frac{(1+7i)(3+4i)}{3^2-(4i{)}^2}$

$=\frac{3+4i+21i+28i^2}{9-16i^2}$

$=\frac{3+25i-28}{9+16}$

$=\frac{-25+25i}{25}$

$=-1+i$

$arg\left(z\right)=\theta ={\tan }^{-1}\left(\frac{1}{-1}\right)$

$={\tan }^{-1}(-1)$

$z$ lies in the second quadrant, so principal value of $\theta =\frac{3\pi }{4}$

So, the given statement is false.