Question

If $z=\sqrt{20i-21}+\sqrt{21+20i}$ then principal value of arg z can not be

Answer 1

$\sqrt{20i+21}=\sqrt{21+20i}=\sqrt{25-4+20i}$

$=\sqrt{{\left(5\right)}^2+{\left(2i\right)}^2+2\times 5\times 2i}$

$=\sqrt{{(5+2i)}^2}$

$\Rightarrow 20i+21={(5+2i)}^2$

$\sqrt{20i-21}=\sqrt{4-25+20i}$

$=\sqrt{{\left(2\right)}^2+{\left(5i\right)}^2+20i}$

$=\sqrt{{(2-5i)}^2}$

$\Rightarrow 20i-21={(2-5i)}^2$

$\Rightarrow z=\pm (5+2i)\pm (2-5i)$

$\Rightarrow z=5+2i+25i,5+2i-2-5i,-5-2i+2+5i,-5-2i-2-5i$

$\therefore z=7+7i,3-3i,-3+3i,-7-7i$

$=7(1+i),3(1-i)3(-1+i),7(-1-i)$

$\Rightarrow arg\left(z\right)=\frac{\pi }{4},\frac{-\pi }{4},\frac{3\pi }{4},\frac{-3\pi }{4}$

$\therefore \arg z$ cannot be more than $\frac{3\pi }{4}$ and less than $\frac{-\pi }{4}$