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Question Number 199290 by ajfour last updated on 31/Oct/23
If x=(√(p+iq))+(√(h+ik)) and (p/q)≠(k/h) then relate p,q,h,k ∈R such that x∈R.
$${If}\:{x}=\sqrt{{p}+{iq}}+\sqrt{{h}+{ik}} \\ $$$${and}\:\:\frac{{p}}{{q}}\neq\frac{{k}}{{h}}\:\:{then}\:{relate}\:{p},{q},{h},{k}\:\in\mathbb{R} \\ $$$${such}\:{that}\:{x}\in\mathbb{R}. \\ $$
Commented by Frix last updated on 31/Oct/23
(√(x+yi))=re^(iθ) with r>0∧−(π/2)<θ≤(π/2) (√(x+yi))∉R ⇒ −(π/2)<θ<0∨0<θ<(π/2) ⇒ (√(p+qi))=a+bi is in the 1^(st) quadrant (√(h+ki))=c+di is in the 4^(th) quadrant (or the other way round) ⇒ for a, b, c, d >0 we get (√(p+qi))=a+bi (√(h+ki))=c−di (√(p+qi))+(√(h+ki))=(a+c)+(b−d)i∈R ⇒ d=b (√(p+qi))=a+bi (√(h+ki))=c−bi p+qi=(a^2 −b^2 )+2abi ⇒ q>0 h+ki=(c^2 −b^2 )−2bci ⇒ k<0 For given p∈R, q∈R^+ a=(√(((√(p^2 +q^2 ))+p)/2)); b=(√(((√(p^2 +q^2 ))−p)/2)) We can find h∈R, k∈R^− 1. For given h∈R c=(√(b^2 +h)); k=−(√(4(h−p)b^2 +q^2 )) 2. For given k∈R^− c=−((ak)/q); h=((a^2 k^2 )/q^2 )−b^2 3. For given r>b^2 c=(√(r−b^2 )); h=r−2b^2 ; k=−2b(√(r−b^2 )) 4. For given −π<θ<0 c=−bcot (θ/2) ; h=((2b^2 cos θ)/(1−cos θ)); k=2b^2 cot (θ/2) I′m not sure what kind of function you want...
$$\sqrt{{x}+{y}\mathrm{i}}={r}\mathrm{e}^{\mathrm{i}\theta} \:\mathrm{with}\:{r}>\mathrm{0}\wedge−\frac{\pi}{\mathrm{2}}<\theta\leqslant\frac{\pi}{\mathrm{2}} \\ $$$$\sqrt{{x}+{y}\mathrm{i}}\notin\mathbb{R}\:\Rightarrow\:−\frac{\pi}{\mathrm{2}}<\theta<\mathrm{0}\vee\mathrm{0}<\theta<\frac{\pi}{\mathrm{2}} \\ $$$$\Rightarrow \\ $$$$\sqrt{{p}+{q}\mathrm{i}}={a}+{b}\mathrm{i}\:\mathrm{is}\:\mathrm{in}\:\mathrm{the}\:\mathrm{1}^{\mathrm{st}} \:\mathrm{quadrant} \\ $$$$\sqrt{{h}+{k}\mathrm{i}}={c}+{d}\mathrm{i}\:\mathrm{is}\:\mathrm{in}\:\mathrm{the}\:\mathrm{4}^{\mathrm{th}} \:\mathrm{quadrant} \\ $$$$\left(\mathrm{or}\:\mathrm{the}\:\mathrm{other}\:\mathrm{way}\:\mathrm{round}\right) \\ $$$$\Rightarrow\:\mathrm{for}\:{a},\:{b},\:{c},\:{d}\:>\mathrm{0}\:\mathrm{we}\:\mathrm{get} \\ $$$$\sqrt{{p}+{q}\mathrm{i}}={a}+{b}\mathrm{i} \\ $$$$\sqrt{{h}+{k}\mathrm{i}}={c}−{d}\mathrm{i} \\ $$$$\sqrt{{p}+{q}\mathrm{i}}+\sqrt{{h}+{k}\mathrm{i}}=\left({a}+{c}\right)+\left({b}−{d}\right)\mathrm{i}\in\mathbb{R}\:\Rightarrow\:{d}={b} \\ $$$$\sqrt{{p}+{q}\mathrm{i}}={a}+{b}\mathrm{i} \\ $$$$\sqrt{{h}+{k}\mathrm{i}}={c}−{b}\mathrm{i} \\ $$$${p}+{q}\mathrm{i}=\left({a}^{\mathrm{2}} −{b}^{\mathrm{2}} \right)+\mathrm{2}{ab}\mathrm{i}\:\Rightarrow\:{q}>\mathrm{0} \\ $$$${h}+{k}\mathrm{i}=\left({c}^{\mathrm{2}} −{b}^{\mathrm{2}} \right)−\mathrm{2}{bc}\mathrm{i}\:\Rightarrow\:{k}<\mathrm{0} \\ $$$$\mathrm{For}\:\mathrm{given}\:{p}\in\mathbb{R},\:{q}\in\mathbb{R}^{+} \\ $$$${a}=\sqrt{\frac{\sqrt{{p}^{\mathrm{2}} +{q}^{\mathrm{2}} }+{p}}{\mathrm{2}}};\:{b}=\sqrt{\frac{\sqrt{{p}^{\mathrm{2}} +{q}^{\mathrm{2}} }−{p}}{\mathrm{2}}} \\ $$$$\mathrm{We}\:\mathrm{can}\:\mathrm{find}\:{h}\in\mathbb{R},\:{k}\in\mathbb{R}^{−} \\ $$$$\mathrm{1}.\:\mathrm{For}\:\mathrm{given}\:{h}\in\mathbb{R} \\ $$$${c}=\sqrt{{b}^{\mathrm{2}} +{h}};\:{k}=−\sqrt{\mathrm{4}\left({h}−{p}\right){b}^{\mathrm{2}} +{q}^{\mathrm{2}} } \\ $$$$\mathrm{2}.\:\mathrm{For}\:\mathrm{given}\:{k}\in\mathbb{R}^{−} \\ $$$${c}=−\frac{{ak}}{{q}};\:{h}=\frac{{a}^{\mathrm{2}} {k}^{\mathrm{2}} }{{q}^{\mathrm{2}} }−{b}^{\mathrm{2}} \\ $$$$\mathrm{3}.\:\mathrm{For}\:\mathrm{given}\:{r}>{b}^{\mathrm{2}} \\ $$$${c}=\sqrt{{r}−{b}^{\mathrm{2}} };\:{h}={r}−\mathrm{2}{b}^{\mathrm{2}} ;\:{k}=−\mathrm{2}{b}\sqrt{{r}−{b}^{\mathrm{2}} } \\ $$$$\mathrm{4}.\:\mathrm{For}\:\mathrm{given}\:−\pi<\theta<\mathrm{0} \\ $$$${c}=−{b}\mathrm{cot}\:\frac{\theta}{\mathrm{2}}\:;\:{h}=\frac{\mathrm{2}{b}^{\mathrm{2}} \mathrm{cos}\:\theta}{\mathrm{1}−\mathrm{cos}\:\theta};\:{k}=\mathrm{2}{b}^{\mathrm{2}} \mathrm{cot}\:\frac{\theta}{\mathrm{2}} \\ $$$$\mathrm{I}'\mathrm{m}\:\mathrm{not}\:\mathrm{sure}\:\mathrm{what}\:\mathrm{kind}\:\mathrm{of}\:\mathrm{function}\:\mathrm{you} \\ $$$$\mathrm{want}… \\ $$
Answered by AST last updated on 31/Oct/23
(√(p+iq))=a+ib ⇒p+iq=a^2 −b^2 +2iab (√(h+ik))=c+di⇒(√(p+iq))+(√(h+ik))=(a+c)+(b+d)i ⇒b=−d ⇒(√(h+ik))=c−ib⇒h+ik=c^2 −b^2 −2ibc ⇒h=(c^2 −b^2 );k=−2bc;p=a^2 −b^2 ;q=2ab
$$\sqrt{{p}+{iq}}={a}+{ib} \\ $$$$\Rightarrow{p}+{iq}={a}^{\mathrm{2}} −{b}^{\mathrm{2}} +\mathrm{2}{iab} \\ $$$$\sqrt{{h}+{ik}}={c}+{di}\Rightarrow\sqrt{{p}+{iq}}+\sqrt{{h}+{ik}}=\left({a}+{c}\right)+\left({b}+{d}\right){i} \\ $$$$\Rightarrow{b}=−{d} \\ $$$$\Rightarrow\sqrt{{h}+{ik}}={c}−{ib}\Rightarrow{h}+{ik}={c}^{\mathrm{2}} −{b}^{\mathrm{2}} −\mathrm{2}{ibc} \\ $$$$\Rightarrow{h}=\left({c}^{\mathrm{2}} −{b}^{\mathrm{2}} \right);{k}=−\mathrm{2}{bc};{p}={a}^{\mathrm{2}} −{b}^{\mathrm{2}} ;{q}=\mathrm{2}{ab} \\ $$
Commented by ajfour last updated on 31/Oct/23
there exists a zuch relation tough to arrive at. I want f(p,q,h,k)=0. Thank you still.
$${there}\:{exists}\:{a}\:{zuch}\:{relation} \\ $$$${tough}\:{to}\:{arrive}\:{at}.\:{I}\:{want} \\ $$$${f}\left({p},{q},{h},{k}\right)=\mathrm{0}.\:{Thank}\:{you}\:{still}. \\ $$
Answered by mr W last updated on 31/Oct/23
(√(p+qi))=[(√(p^2 +q^2 ))((p/( (√(p^2 +q^2 ))))+(q/( (√(p^2 +q^2 ))))i)]^(1/2) =[(√(p^2 +q^2 )) e^((tan^(−1) (q/p))i) ]^(1/2) =(p^2 +q^2 )^(1/4) e^((1/2)(tan^(−1) (q/p))i) similarly (√(h+ki))=(h^2 +k^2 )^(1/4) e^((1/2)(tan^(−1) (k/h))i) x=(√(p+qi))+(√(h+ki))=(p^2 +q^2 )^(1/4) e^((1/2)(tan^(−1) (q/p))i) +(h^2 +k^2 )^(1/4) e^((1/2)(tan^(−1) (k/h))i) such that x∈R, ⇒(p^2 +q^2 )^(1/4) sin ((tan^(−1) (q/p))/2)+(h^2 +k^2 )^(1/4) sin ((tan^(−1) (k/h))/2)=0
$$\sqrt{{p}+{qi}}=\left[\sqrt{{p}^{\mathrm{2}} +{q}^{\mathrm{2}} }\left(\frac{{p}}{\:\sqrt{{p}^{\mathrm{2}} +{q}^{\mathrm{2}} }}+\frac{{q}}{\:\sqrt{{p}^{\mathrm{2}} +{q}^{\mathrm{2}} }}{i}\right)\right]^{\frac{\mathrm{1}}{\mathrm{2}}} \\ $$$$=\left[\sqrt{{p}^{\mathrm{2}} +{q}^{\mathrm{2}} }\:{e}^{\left(\mathrm{tan}^{−\mathrm{1}} \frac{{q}}{{p}}\right){i}} \right]^{\frac{\mathrm{1}}{\mathrm{2}}} \\ $$$$=\left({p}^{\mathrm{2}} +{q}^{\mathrm{2}} \right)^{\frac{\mathrm{1}}{\mathrm{4}}} {e}^{\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{tan}^{−\mathrm{1}} \frac{{q}}{{p}}\right){i}} \\ $$$${similarly} \\ $$$$\sqrt{{h}+{ki}}=\left({h}^{\mathrm{2}} +{k}^{\mathrm{2}} \right)^{\frac{\mathrm{1}}{\mathrm{4}}} {e}^{\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{tan}^{−\mathrm{1}} \frac{{k}}{{h}}\right){i}} \\ $$$${x}=\sqrt{{p}+{qi}}+\sqrt{{h}+{ki}}=\left({p}^{\mathrm{2}} +{q}^{\mathrm{2}} \right)^{\frac{\mathrm{1}}{\mathrm{4}}} {e}^{\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{tan}^{−\mathrm{1}} \frac{{q}}{{p}}\right){i}} +\left({h}^{\mathrm{2}} +{k}^{\mathrm{2}} \right)^{\frac{\mathrm{1}}{\mathrm{4}}} {e}^{\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{tan}^{−\mathrm{1}} \frac{{k}}{{h}}\right){i}} \\ $$$${such}\:{that}\:{x}\in{R}, \\ $$$$\Rightarrow\left({p}^{\mathrm{2}} +{q}^{\mathrm{2}} \right)^{\frac{\mathrm{1}}{\mathrm{4}}} \:\mathrm{sin}\:\frac{\mathrm{tan}^{−\mathrm{1}} \frac{{q}}{{p}}}{\mathrm{2}}+\left({h}^{\mathrm{2}} +{k}^{\mathrm{2}} \right)^{\frac{\mathrm{1}}{\mathrm{4}}} \mathrm{sin}\:\frac{\mathrm{tan}^{−\mathrm{1}} \frac{{k}}{{h}}}{\mathrm{2}}=\mathrm{0} \\ $$
Commented by ajfour last updated on 31/Oct/23
great way, thanks immensely sir!
$${great}\:{way},\:{thanks}\:{immensely}\:{sir}! \\ $$
Answered by MM42 last updated on 31/Oct/23
p=rcosa & q=rsina & r=(√(p^2 +q^2 )) h=r′cosb & k=r′sinb & r′=(√(h^2 +k^2 )) ⇒x=(√r)e^(i(a/2)) +(√(r′))e^(i(b/2)) ⇒x^2 =re^(ia) +r^′ e^(ib) +2(√(rr′))e^(i(((a+b)/2))) =r(cosa+isina)+r′(cosb+isinb)+2(√(rr′))(cos(((a+b)/2))+isin(((a+b)/2))) x∈R⇒x^2 ∈R ⇒rsina+r′sinb+2(√(rr′))sin(((a+b)/2))=0 ⇒rsina+r′sinb+2(√(rr′))×(√((1−cos(a+b))/2))=0 ⇒p+q+2(√((rr′−(rsina)(r′cosb)+(r′sinb)(rcosa))/2))=0 ⇒p+q+2(√((rr′−qh+pk)/2))=0 ⇒p+q+(√(2((√((p^2 +q^2 )(h^2 +k^2 )))+pk−qh)))=0
$$\:{p}={rcosa}\:\&\:{q}={rsina}\:\:\&\:{r}=\sqrt{{p}^{\mathrm{2}} +{q}^{\mathrm{2}} } \\ $$$$\:{h}={r}'{cosb}\:\:\&\:\:{k}={r}'{sinb}\:\:\:\&\:\:\:{r}'=\sqrt{{h}^{\mathrm{2}} +{k}^{\mathrm{2}} } \\ $$$$\Rightarrow{x}=\sqrt{{r}}{e}^{{i}\frac{{a}}{\mathrm{2}}} +\sqrt{{r}'}{e}^{{i}\frac{{b}}{\mathrm{2}}} \\ $$$$\Rightarrow{x}^{\mathrm{2}} ={re}^{{ia}} +{r}^{'} {e}^{{ib}} +\mathrm{2}\sqrt{{rr}'}{e}^{{i}\left(\frac{{a}+{b}}{\mathrm{2}}\right)} \\ $$$$={r}\left({cosa}+{isina}\right)+{r}'\left({cosb}+{isinb}\right)+\mathrm{2}\sqrt{{rr}'}\left({cos}\left(\frac{{a}+{b}}{\mathrm{2}}\right)+{isin}\left(\frac{{a}+{b}}{\mathrm{2}}\right)\right) \\ $$$${x}\in\mathbb{R}\Rightarrow{x}^{\mathrm{2}} \in\mathbb{R}\:\Rightarrow{rsina}+{r}'{sinb}+\mathrm{2}\sqrt{{rr}'}{sin}\left(\frac{{a}+{b}}{\mathrm{2}}\right)=\mathrm{0} \\ $$$$\Rightarrow{rsina}+{r}'{sinb}+\mathrm{2}\sqrt{{rr}'}×\sqrt{\frac{\mathrm{1}−{cos}\left({a}+{b}\right)}{\mathrm{2}}}=\mathrm{0} \\ $$$$\Rightarrow{p}+{q}+\mathrm{2}\sqrt{\frac{{rr}'−\left({rsina}\right)\left({r}'{cosb}\right)+\left({r}'{sinb}\right)\left({rcosa}\right)}{\mathrm{2}}}=\mathrm{0} \\ $$$$\Rightarrow{p}+{q}+\mathrm{2}\sqrt{\frac{{rr}'−{qh}+{pk}}{\mathrm{2}}}=\mathrm{0} \\ $$$$\Rightarrow{p}+{q}+\sqrt{\mathrm{2}\left(\sqrt{\left({p}^{\mathrm{2}} +{q}^{\mathrm{2}} \right)\left({h}^{\mathrm{2}} +{k}^{\mathrm{2}} \right)}+{pk}−{qh}\right)}=\mathrm{0} \\ $$$$ \\ $$
Commented by mr W last updated on 31/Oct/23
can we say x∈R ⇔ x^2 ∈R ? i think no. example: x^2 =−2 ∈R, but x=±(√2)i ∉R.
$${can}\:{we}\:{say} \\ $$$${x}\in{R}\:\Leftrightarrow\:{x}^{\mathrm{2}} \in{R}\:? \\ $$$${i}\:{think}\:{no}.\:{example}:\:{x}^{\mathrm{2}} =−\mathrm{2}\:\in{R},\:{but} \\ $$$${x}=\pm\sqrt{\mathrm{2}}{i}\:\notin{R}. \\ $$
Commented by MM42 last updated on 31/Oct/23
Sir W if “ x∈R”⇒x^2 ∈R but “x^2 ∈R”⇏x∈R i did not get such a result “ x^2 ∈R⇒x∈R”
$${Sir}\:{W} \\ $$$${if}\:“\:{x}\in\mathbb{R}''\Rightarrow{x}^{\mathrm{2}} \in\mathbb{R} \\ $$$${but}\:\:“{x}^{\mathrm{2}} \in\mathbb{R}''\nRightarrow{x}\in\mathbb{R} \\ $$$${i}\:{did}\:{not}\:{get}\:{such}\:{a}\:{result} \\ $$$$“\:{x}^{\mathrm{2}} \in\mathbb{R}\Rightarrow{x}\in\mathbb{R}'' \\ $$
Commented by MM42 last updated on 31/Oct/23
yes result of the last line was incorrect. the result is the same line before the end. but “if z^2 =0⇒z=0 ”
$${yes}\:{result}\:{of}\:\:{the}\:{last}\:{line}\:{was}\:{incorrect}. \\ $$$${the}\:{result}\:{is}\:{the}\:{same}\:{line}\:{before} \\ $$$${the}\:{end}.\: \\ $$$${but}\:\:“{if}\:\:\:{z}^{\mathrm{2}} =\mathrm{0}\Rightarrow{z}=\mathrm{0}\:'' \\ $$
Commented by mr W last updated on 31/Oct/23
with (√(p+q+(√(2(√((p^2 +q^2 )(h^2 +k^2 )))+pk−qh))))=0 ✓ we ensure that x^2 ∈R. but are we also sure that x∈R?
$${with} \\ $$$$\sqrt{{p}+{q}+\sqrt{\mathrm{2}\sqrt{\left({p}^{\mathrm{2}} +{q}^{\mathrm{2}} \right)\left({h}^{\mathrm{2}} +{k}^{\mathrm{2}} \right)}+{pk}−{qh}}}=\mathrm{0}\:\checkmark \\ $$$${we}\:{ensure}\:{that}\:{x}^{\mathrm{2}} \in\mathbb{R}.\:{but}\:{are}\:{we}\:{also} \\ $$$${sure}\:{that}\:{x}\in\mathbb{R}? \\ $$
Answered by Mathspace last updated on 31/Oct/23
p+iq=(√(p^2 +q^2 ))e^(iarctan((q/p))) and (√(p+iq))=(p^2 +q^2 )^(1/4) e^((1/2)iarctan((q/p))) also (√(h+ik))=(h^2 +k^2 )^(1/4) e^((1/2)iarctan((k/h))) ⇒x=^4 (√(p^2 +q^2 )){cos((1/2)arctan((q/p)) +isin((1/2)arctan((q/p))} +^4 (√(h^2 +k^2 )){cos((1/2)arctan((k/h))) +isin((1/2)arctan((k/h))} and x∈R ⇒Im(....)=0 ⇔ (^4 (√(p^2 +q^2 )))sin((1/2)arctan((q/p))) +^4 (√(h^2 +k^2 ))sin((1/2)arctan((k/h)))=0⇒ ((sin((1/2)arctan((q/p)))/(sin((1/2)arctan((k/h))))=−^4 (√((h^2 +k^2 )/(p^2 +q^2 )))
$${p}+{iq}=\sqrt{{p}^{\mathrm{2}} +{q}^{\mathrm{2}} }{e}^{{iarctan}\left(\frac{{q}}{{p}}\right)} \\ $$$${and}\:\sqrt{{p}+{iq}}=\left({p}^{\mathrm{2}} +{q}^{\mathrm{2}} \right)^{\frac{\mathrm{1}}{\mathrm{4}}} {e}^{\frac{\mathrm{1}}{\mathrm{2}}{iarctan}\left(\frac{{q}}{{p}}\right)} \\ $$$${also} \\ $$$$\sqrt{{h}+{ik}}=\left({h}^{\mathrm{2}} +{k}^{\mathrm{2}} \right)^{\frac{\mathrm{1}}{\mathrm{4}}} \:{e}^{\frac{\mathrm{1}}{\mathrm{2}}{iarctan}\left(\frac{{k}}{{h}}\right)} \\ $$$$\Rightarrow{x}=^{\mathrm{4}} \sqrt{{p}^{\mathrm{2}} +{q}^{\mathrm{2}} }\left\{{cos}\left(\frac{\mathrm{1}}{\mathrm{2}}{arctan}\left(\frac{{q}}{{p}}\right)\right.\right. \\ $$$$+{isin}\left(\frac{\mathrm{1}}{\mathrm{2}}{arctan}\left(\frac{{q}}{{p}}\right)\right\} \\ $$$$+^{\mathrm{4}} \sqrt{{h}^{\mathrm{2}} +{k}^{\mathrm{2}} }\left\{{cos}\left(\frac{\mathrm{1}}{\mathrm{2}}{arctan}\left(\frac{{k}}{{h}}\right)\right)\right. \\ $$$$+{isin}\left(\frac{\mathrm{1}}{\mathrm{2}}{arctan}\left(\frac{{k}}{{h}}\right)\right\} \\ $$$${and}\:{x}\in{R}\:\Rightarrow{Im}\left(….\right)=\mathrm{0}\:\Leftrightarrow \\ $$$$\left(^{\mathrm{4}} \sqrt{{p}^{\mathrm{2}} +{q}^{\mathrm{2}} }\right){sin}\left(\frac{\mathrm{1}}{\mathrm{2}}{arctan}\left(\frac{{q}}{{p}}\right)\right) \\ $$$$+^{\mathrm{4}} \sqrt{{h}^{\mathrm{2}} +{k}^{\mathrm{2}} }{sin}\left(\frac{\mathrm{1}}{\mathrm{2}}{arctan}\left(\frac{{k}}{{h}}\right)\right)=\mathrm{0}\Rightarrow \\ $$$$\frac{{sin}\left(\frac{\mathrm{1}}{\mathrm{2}}{arctan}\left(\frac{{q}}{{p}}\right)\right.}{{sin}\left(\frac{\mathrm{1}}{\mathrm{2}}{arctan}\left(\frac{{k}}{{h}}\right)\right.}=−^{\mathrm{4}} \sqrt{\frac{{h}^{\mathrm{2}} +{k}^{\mathrm{2}} }{{p}^{\mathrm{2}} +{q}^{\mathrm{2}} }} \\ $$
Answered by ajfour last updated on 01/Nov/23
x^2 =p+h+i(q+k)+2(√w) (√w)=(√((p+iq)(h+ik)))=s−i(q+k) ⇒ ph−qk+i(pk+hq) = s^2 −(q+k)^2 −2is(q+k) ph−qk=s^2 −(q+k)^2 2s= ((pk+hq)/(k+q)) 4(ph−qk)+4(q+k)^2 =(((pk+hq)/(k+q)))^2
$${x}^{\mathrm{2}} ={p}+{h}+{i}\left({q}+{k}\right)+\mathrm{2}\sqrt{{w}} \\ $$$$\sqrt{{w}}=\sqrt{\left({p}+{iq}\right)\left({h}+{ik}\right)}={s}−{i}\left({q}+{k}\right) \\ $$$$\Rightarrow\:{ph}−{qk}+{i}\left({pk}+{hq}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:=\:{s}^{\mathrm{2}} −\left({q}+{k}\right)^{\mathrm{2}} −\mathrm{2}{is}\left({q}+{k}\right) \\ $$$${ph}−{qk}={s}^{\mathrm{2}} −\left({q}+{k}\right)^{\mathrm{2}} \\ $$$$\mathrm{2}{s}=\:\frac{{pk}+{hq}}{{k}+{q}} \\ $$$$\mathrm{4}\left({ph}−{qk}\right)+\mathrm{4}\left({q}+{k}\right)^{\mathrm{2}} =\left(\frac{{pk}+{hq}}{{k}+{q}}\right)^{\mathrm{2}} \\ $$

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