Question

Find (i) the square root of $7+24i$ (ii) $\sqrt{i}+\sqrt{-1}$

Answer 1

1. $\sqrt{7+24i}=a+ib$ (let)

$7+24i={(a+ib)}^2$

$a^2-b^2=7,ab=12$

$ab=12$

$b=\frac{12}{a}$

$a^2-\frac{144}{a^2}=7$

$a^4-144=7a^2$

$a^4-7a^2-144=0$

$a^4-16a^2+9a^2-144=0$

$a^2(a^2-16)+9(a^2-16)=0$

$(a^2-9)(a^2-16)=0$

$a=\pm 3,\pm 4$

$a^2-b^2=7$

$a=\pm 3,b=\pm \sqrt{2};a=\pm 4,b=\pm 3$

$\sqrt{7+24i}=\pm (3+\sqrt{2}i)$ or $\pm (4+3i)$

2. $\sqrt{i}=a+ib$ (let)

$i=a^2-b^2+2abi$

$a^2=b^2,ab=\frac{1}{2}$

$a=\frac{1}{2b}$

$\frac{1}{4b^2}=b^2$

$b=\pm \frac{1}{\sqrt{2}},a=\pm \frac{1}{\sqrt{2}}$

$\sqrt{i}=\pm \frac{1}{\sqrt{2}}(1+i)$

$\sqrt{-1}=i$

$\sqrt{i}+\sqrt{-1}=\pm \frac{1}{\sqrt{2}}(1+i)+i$