By taking sqrt{a pm i b}=x pm i y, x0,if we | vedclass

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(A) Let $z = \frac{\sqrt{21+12 \sqrt{2} i}}{\sqrt{21-12 \sqrt{2} i}}$.Note that $21+12 \sqrt{2} i = (2 \sqrt{2} + 3i)^2$ because $(2 \sqrt{2})^2 - 3^2

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$\frac{4 \sqrt{2}}{7}$

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(A) Let $z = \frac{\sqrt{21+12 \sqrt{2} i}}{\sqrt{21-12 \sqrt{2} i}}$.Note that $21+12 \sqrt{2} i = (2 \sqrt{2} + 3i)^2$ because $(2 \sqrt{2})^2 - 3^2 + 2(2 \sqrt{2})(3)i = 8 - 9 + 12 \sqrt{2} i = -1 + 12 \sqrt{2} i$. This is not correct.Let us find the square root of $21+12 \sqrt{2} i$. Let $(x+iy)^2 = 21+12 \sqrt{2} i$.$x^2-y^2 = 21$ and $2xy = 12 \sqrt{2} \implies xy = 6 \sqrt{2}$.$(x^2+y^2)^2 = (x^2-y^2)^2 + (2xy)^2 = 21^2 + (12 \sqrt{2})^2 = 441 + 288 = 729$.$x^2+y^2 = 27$.Adding $x^2-y^2=21$ and $x^2+y^2=27$,we get $2x^2 = 48 \implies x^2 = 24 \implies x = 2 \sqrt{6}$.$2y^2 = 6 \implies y^2 = 3 \implies y = \sqrt{3}$.So,$\sqrt{21+12 \sqrt{2} i} = 2 \sqrt{6} + i \sqrt{3}$.Similarly,$\sqrt{21-12 \sqrt{2} i} = 2 \sqrt{6} - i \sqrt{3}$.Then $z = \frac{2 \sqrt{6} + i \sqrt{3}}{2 \sqrt{6} - i \sqrt{3}} = \frac{(2 \sqrt{6} + i \sqrt{3})^2}{(2 \sqrt{6})^2 + (\sqrt{3})^2} = \frac{24 - 3 + 4 \sqrt{18} i}{24 + 3} = \frac{21 + 12 \sqrt{2} i}{27} = \frac{21}{27} + i \frac{12 \sqrt{2}}{27} = \frac{7}{9} + i \frac{4 \sqrt{2}}{9}$.Here $a = \frac{7}{9}$ and $b = \frac{4 \sqrt{2}}{9}$.Thus,$\frac{b}{a} = \frac{4 \sqrt{2} / 9}{7 / 9} = \frac{4 \sqrt{2}}{7}$.

By taking $\sqrt{a \pm i b}=x \pm i y, x>0$,if we get $\frac{\sqrt{21+12 \sqrt{2} i}}{\sqrt{21-12 \sqrt{2} i}}=a+i b$,then $\frac{b}{a}=$

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