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Question Number 199290 by ajfour last updated on 31/Oct/23

$$Ifx=\sqrt{p+iq}+\sqrt{h+ik}$$$$and\frac{p}{q}\ne \frac{k}{h}thenrelatep,q,h,k\in \mathbb{R}$$$$suchthatx\in \mathbb{R}.$$

Commented by Frix last updated on 31/Oct/23

$$\sqrt{x+y\mathrm{i}}=r{\mathrm{e}}^{\mathrm{i}\theta }\mathrm{with}r>0\wedge -\frac{\pi }{2}<\theta ⩽\frac{\pi }{2}$$$$\sqrt{x+y\mathrm{i}}\notin \mathbb{R}\Rightarrow -\frac{\pi }{2}<\theta <0\vee 0<\theta <\frac{\pi }{2}$$$$\Rightarrow$$$$\sqrt{p+q\mathrm{i}}=a+b\mathrm{i}\mathrm{is}\mathrm{in}\mathrm{the}{1}^{\mathrm{st}}\mathrm{quadrant}$$$$\sqrt{h+k\mathrm{i}}=c+d\mathrm{i}\mathrm{is}\mathrm{in}\mathrm{the}{4}^{\mathrm{th}}\mathrm{quadrant}$$$$\left(\mathrm{or}\mathrm{the}\mathrm{other}\mathrm{way}\mathrm{round}\right)$$$$\Rightarrow \mathrm{for}a,b,c,d>0\mathrm{we}\mathrm{get}$$$$\sqrt{p+q\mathrm{i}}=a+b\mathrm{i}$$$$\sqrt{h+k\mathrm{i}}=c-d\mathrm{i}$$$$\sqrt{p+q\mathrm{i}}+\sqrt{h+k\mathrm{i}}=(a+c)+(b-d)\mathrm{i}\in \mathbb{R}\Rightarrow d=b$$$$\sqrt{p+q\mathrm{i}}=a+b\mathrm{i}$$$$\sqrt{h+k\mathrm{i}}=c-b\mathrm{i}$$$$p+q\mathrm{i}=(a^2-b^2)+2ab\mathrm{i}\Rightarrow q>0$$$$h+k\mathrm{i}=(c^2-b^2)-2bc\mathrm{i}\Rightarrow k<0$$$$\mathrm{For}\mathrm{given}p\in \mathbb{R},q\in {\mathbb{R}}^{+}$$$$a=\sqrt{\frac{\sqrt{p^2+q^2}+p}{2}};b=\sqrt{\frac{\sqrt{p^2+q^2}-p}{2}}$$$$\mathrm{We}\mathrm{can}\mathrm{find}h\in \mathbb{R},k\in {\mathbb{R}}^{-}$$$$1.\mathrm{For}\mathrm{given}h\in \mathbb{R}$$$$c=\sqrt{b^2+h};k=-\sqrt{4(h-p)b^2+q^2}$$$$2.\mathrm{For}\mathrm{given}k\in {\mathbb{R}}^{-}$$$$c=-\frac{ak}{q};h=\frac{a^2{k}^2}{q^2}-b^2$$$$3.\mathrm{For}\mathrm{given}r>b^2$$$$c=\sqrt{r-b^2};h=r-2b^2;k=-2b\sqrt{r-b^2}$$$$4.\mathrm{For}\mathrm{given}-\pi <\theta <0$$$$c=-b\cot \frac{\theta }{2};h=\frac{2b^2\cos \theta }{1-\cos \theta };k=2b^2\cot \frac{\theta }{2}$$$${\mathrm{I}}^{\prime }\mathrm{m}\mathrm{not}\mathrm{sure}\mathrm{what}\mathrm{kind}\mathrm{of}\mathrm{function}\mathrm{you}$$$$\mathrm{want}...$$

Answered by mr W last updated on 31/Oct/23

$$\sqrt{p+qi}={\left[\sqrt{p^2+q^2}(\frac{p}{\sqrt{p^2+q^2}}+\frac{q}{\sqrt{p^2+q^2}}i)\right]}^{\frac{1}{2}}$$$$={\left[\sqrt{p^2+q^2}{e}^{\left({\tan }^{-1}\frac{q}{p}\right)i}\right]}^{\frac{1}{2}}$$$$={(p^2+q^2)}^{\frac{1}{4}}{e}^{\frac{1}{2}\left({\tan }^{-1}\frac{q}{p}\right)i}$$$$similarly$$$$\sqrt{h+ki}={(h^2+k^2)}^{\frac{1}{4}}{e}^{\frac{1}{2}\left({\tan }^{-1}\frac{k}{h}\right)i}$$$$x=\sqrt{p+qi}+\sqrt{h+ki}={(p^2+q^2)}^{\frac{1}{4}}{e}^{\frac{1}{2}\left({\tan }^{-1}\frac{q}{p}\right)i}+{(h^2+k^2)}^{\frac{1}{4}}{e}^{\frac{1}{2}\left({\tan }^{-1}\frac{k}{h}\right)i}$$$$suchthatx\in R,$$$$\Rightarrow {(p^2+q^2)}^{\frac{1}{4}}\sin \frac{{\tan }^{-1}\frac{q}{p}}{2}+{(h^2+k^2)}^{\frac{1}{4}}\sin \frac{{\tan }^{-1}\frac{k}{h}}{2}=0$$

Commented by ajfour last updated on 31/Oct/23

$$greatway,thanksimmenselysir!$$

Answered by MM42 last updated on 31/Oct/23

$$p=rcosa\mathrm{\&}q=rsina\mathrm{\&}r=\sqrt{p^2+q^2}$$$$h=r^{\prime }cosb\mathrm{\&}k=r^{\prime }sinb\mathrm{\&}{r}^{\prime }=\sqrt{h^2+k^2}$$$$\Rightarrow x=\sqrt{r}{e}^{i\frac{a}{2}}+\sqrt{r^{\prime }}{e}^{i\frac{b}{2}}$$$$\Rightarrow x^2={re}^{ia}+r^{{}^{\prime }}{e}^{ib}+2\sqrt{{rr}^{\prime }}{e}^{i\left(\frac{a+b}{2}\right)}$$$$=r(cosa+isina)+r^{\prime }(cosb+isinb)+2\sqrt{{rr}^{\prime }}(cos\left(\frac{a+b}{2}\right)+isin\left(\frac{a+b}{2}\right))$$$$x\in \mathbb{R}\Rightarrow x^2\in \mathbb{R}\Rightarrow rsina+r^{\prime }sinb+2\sqrt{{rr}^{\prime }}sin\left(\frac{a+b}{2}\right)=0$$$$\Rightarrow rsina+r^{\prime }sinb+2\sqrt{{rr}^{\prime }}\times \sqrt{\frac{1-cos(a+b)}{2}}=0$$$$\Rightarrow p+q+2\sqrt{\frac{{rr}^{\prime }-\left(rsina\right)\left(r^{\prime }cosb\right)+\left(r^{\prime }sinb\right)\left(rcosa\right)}{2}}=0$$$$\Rightarrow p+q+2\sqrt{\frac{{rr}^{\prime }-qh+pk}{2}}=0$$$$\Rightarrow p+q+\sqrt{2(\sqrt{(p^2+q^2)(h^2+k^2)}+pk-qh)}=0$$$$$$

Commented by mr W last updated on 31/Oct/23

$$canwesay$$$$x\in R\iff x^2\in R?$$$$ithinkno.example:x^2=-2\in R,but$$$$x=\pm \sqrt{2}i\notin R.$$

Commented by MM42 last updated on 31/Oct/23

$$SirW$$$$if‘‘x\in {\mathbb{R}}^{″}\Rightarrow x^2\in \mathbb{R}$$$$but‘‘x^2\in {\mathbb{R}}^{″}\nRightarrow x\in \mathbb{R}$$$$ididnotgetsucharesult$$$$‘‘x^2\in \mathbb{R}\Rightarrow x\in {\mathbb{R}}^{″}$$

Commented by MM42 last updated on 31/Oct/23

$$yesresultofthelastlinewasincorrect.$$$$theresultisthesamelinebefore$$$$theend.$$$$but‘‘if{z}^2=0\Rightarrow z=0{}^{″}$$

Commented by mr W last updated on 31/Oct/23

$$with$$$$\sqrt{p+q+\sqrt{2\sqrt{(p^2+q^2)(h^2+k^2)}+pk-qh}}=0✓$$$$weensurethat{x}^2\in \mathbb{R}.butarewealso$$$$surethatx\in \mathbb{R}?$$

Answered by Mathspace last updated on 31/Oct/23

$$p+iq=\sqrt{p^2+q^2}{e}^{iarctan\left(\frac{q}{p}\right)}$$$$and\sqrt{p+iq}={(p^2+q^2)}^{\frac{1}{4}}{e}^{\frac{1}{2}iarctan\left(\frac{q}{p}\right)}$$$$also$$$$\sqrt{h+ik}={(h^2+k^2)}^{\frac{1}{4}}{e}^{\frac{1}{2}iarctan\left(\frac{k}{h}\right)}$$$$\Rightarrow x{=}^4\sqrt{p^2+q^2}\{cos(\frac{1}{2}arctan\left(\frac{q}{p}\right)$$$$+isin\left(\frac{1}{2}arctan\left(\frac{q}{p}\right)\right\}$$$${+}^4\sqrt{h^2+k^2}\{cos\left(\frac{1}{2}arctan\left(\frac{k}{h}\right)\right)$$$$+isin\left(\frac{1}{2}arctan\left(\frac{k}{h}\right)\right\}$$$$andx\in R\Rightarrow Im(....)=0\iff$$$$\left({}^4\sqrt{p^2+q^2}\right)sin\left(\frac{1}{2}arctan\left(\frac{q}{p}\right)\right)$$$${+}^4\sqrt{h^2+k^2}sin\left(\frac{1}{2}arctan\left(\frac{k}{h}\right)\right)=0\Rightarrow$$$$\frac{sin(\frac{1}{2}arctan\left(\frac{q}{p}\right)}{sin(\frac{1}{2}arctan\left(\frac{k}{h}\right)}={-}^4\sqrt{\frac{h^2+k^2}{p^2+q^2}}$$

Answered by ajfour last updated on 01/Nov/23

$$x^2=p+h+i(q+k)+2\sqrt{w}$$$$\sqrt{w}=\sqrt{(p+iq)(h+ik)}=s-i(q+k)$$$$\Rightarrow ph-qk+i(pk+hq)$$$$=s^2-{(q+k)}^2-2is(q+k)$$$$ph-qk=s^2-{(q+k)}^2$$$$2s=\frac{pk+hq}{k+q}$$$$4(ph-qk)+4{(q+k)}^2={\left(\frac{pk+hq}{k+q}\right)}^2$$

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