sqrt{a pm i b}=x pm i y, x0 लेकर,यदि हमें fr | vedclass

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(A) माना $z = \frac{\sqrt{21+12 \sqrt{2} i}}{\sqrt{21-12 \sqrt{2} i}}$.$21+12 \sqrt{2} i$ का वर्गमूल ज्ञात करने पर,$(x+iy)^2 = 21+12 \sqrt{2} i$ लेने

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

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(A) माना $z = \frac{\sqrt{21+12 \sqrt{2} i}}{\sqrt{21-12 \sqrt{2} i}}$.$21+12 \sqrt{2} i$ का वर्गमूल ज्ञात करने पर,$(x+iy)^2 = 21+12 \sqrt{2} i$ लेने पर.$x^2-y^2 = 21$ और $2xy = 12 \sqrt{2} \implies xy = 6 \sqrt{2}$.$(x^2+y^2)^2 = 21^2 + (12 \sqrt{2})^2 = 729 \implies x^2+y^2 = 27$.$x^2 = 24 \implies x = 2 \sqrt{6}$ और $y^2 = 3 \implies y = \sqrt{3}$.अतः,$\sqrt{21+12 \sqrt{2} i} = 2 \sqrt{6} + i \sqrt{3}$ और $\sqrt{21-12 \sqrt{2} i} = 2 \sqrt{6} - i \sqrt{3}$.$z = \frac{2 \sqrt{6} + i \sqrt{3}}{2 \sqrt{6} - i \sqrt{3}} = \frac{(2 \sqrt{6} + i \sqrt{3})^2}{24 + 3} = \frac{21 + 12 \sqrt{2} i}{27} = \frac{7}{9} + i \frac{4 \sqrt{2}}{9}$.यहाँ $a = \frac{7}{9}$ और $b = \frac{4 \sqrt{2}}{9}$.अतः,$\frac{b}{a} = \frac{4 \sqrt{2}}{7}$.

$\sqrt{a \pm i b}=x \pm i y, x>0$ लेकर,यदि हमें $\frac{\sqrt{21+12 \sqrt{2} i}}{\sqrt{21-12 \sqrt{2} i}}=a+i b$ प्राप्त होता है,तो $\frac{b}{a}=$

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