Question and Answers Forum

All Questions      Topic List

Algebra Questions

Previous in All Question      Next in All Question      

Previous in Algebra      Next in Algebra      

Question Number 121127 by Study last updated on 05/Nov/20

$$\sqrt{3+2i}=?$$

Answered by MJS_new last updated on 05/Nov/20

$$3+2\mathrm{i}=\sqrt{13}{\mathrm{e}}^{\mathrm{i}\arctan \frac{2}{3}}$$$$\sqrt{3+2\mathrm{i}}={13}^{1/4}{\mathrm{e}}^{\mathrm{i}\frac{\arctan \frac{2}{3}}{2}}$$$$\mathrm{or}$$$$\sqrt{a+b\mathrm{i}}=\frac{\sqrt{2(a+\sqrt{a^2+b^2})}}{2}+\mathrm{sign}b\frac{\sqrt{2(-a+\sqrt{a^2+b^2})}}{2}\mathrm{i}$$$$\Rightarrow$$$$\sqrt{3+2\mathrm{i}}=\frac{\sqrt{6+2\sqrt{13}}}{2}+\frac{\sqrt{-6+2\sqrt{13}}}{2}\mathrm{i}$$

Answered by Dwaipayan Shikari last updated on 05/Nov/20

$$Generally$$$$\sqrt{x+i\sqrt{y}}=\sqrt{a}+i\sqrt{b}$$$$x+i\sqrt{y}=a-b+2i\sqrt{ab}$$$$x=a-b,y=4ab$$$$a+b=\sqrt{x^2+y}$$$$a=\frac{x+\sqrt{x^2+y}}{2},b=\frac{x-\sqrt{x^2+y}}{2}$$$$\sqrt{x+i\sqrt{y}}=\sqrt{\frac{x+\sqrt{x^2+y}}{2}}+i\sqrt{\frac{x-\sqrt{x^2+y}}{2}}$$$$\sqrt{3+i\sqrt{4}}=\sqrt{\frac{3+\sqrt{13}}{2}}+i\sqrt{\frac{3-\sqrt{13}}{2}}$$

Answered by floor(10²Eta[1]) last updated on 05/Nov/20

$$\sqrt{3+2\mathrm{i}}=\mathrm{a}+\mathrm{bi},\mathrm{a},\mathrm{b}\in \mathbb{R}$$$$3+2\mathrm{i}={\mathrm{a}}^2-{\mathrm{b}}^2+2\mathrm{abi}$$$$\Rightarrow {\mathrm{a}}^2-{\mathrm{b}}^2=3$$$$2\mathrm{ab}=2\therefore \mathrm{ab}=1$$$${\mathrm{b}}^4+{3\mathrm{b}}^2-1=0$$$${\mathrm{b}}^2=\frac{-3\pm \sqrt{13}}{2}\Rightarrow \mathrm{b}=\pm \sqrt{\frac{\sqrt{13}-3}{2}}$$$$\Rightarrow \mathrm{a}=\pm \sqrt{\frac{\sqrt{13}+3}{2}}$$$$\sqrt{3+2\mathrm{i}}=\sqrt{\frac{\sqrt{13}+3}{2}}+\mathrm{i}\sqrt{\frac{\sqrt{13}-3}{2}}$$$$\sqrt{3+2\mathrm{i}}=-\sqrt{\frac{\sqrt{13}+3}{2}}-\mathrm{i}\sqrt{\frac{\sqrt{13}-3}{2}}$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com