Question

$3-4i$ equals

• A. $5e^{i\{\pi -{\tan }^{-1}(3/4\left)\right\}\}}$
• B. $5e^{-i\{\pi -{\tan }^{-1}(4/3\left)\right\}\}}$
• C. $5e^{i\{\pi -{\tan }^{-1}(4/3\left)\right\}\}}$
• D. $e^{i\{\pi -{\tan }^{-1}(3/4\left)\right\}\}}$

Answer 1

Answer: (B)

$5e^{-i\{\pi -{\tan }^{-1}(4/3\left)\right\}\}}$

To find: $3-4i$

Solution:

Let $z=3-4i$

$\left|z\right|=\sqrt{3^2+4^2}=5\{|a-bi|=\sqrt{a^2+b^2}\}$

$z=3-4i=r{e}^{i\theta }$

$\theta =arg\&r=\left|z\right|$

$\theta ={\tan }^{-1}\frac{b}{a}={\tan }^{-1}\left(\frac{-4}{3}\right)=\pi -{\tan }^{-1}\left(\frac{4}{3}\right)...\{{\tan }^{-1}(-\theta )=\pi -{\tan }^{-1}\left(\theta \right)\}$

Here, $\theta$ lies in the fourth quadrant.

Argument$=-a=-|\pi -{\tan }^{-1}\left(\frac{4}{3}\right)|$

$\therefore z=r{e}^{i\theta }$

$z=5e^{-i[\pi -{\tan }^{-1}\left(\frac{4}{3}\right)]}$