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Question Number 558 by 112358 last updated on 26/Jan/15

Determine the complex number z such that z^2 =tanx+icosx, in the form z=p+ik, p,k∈R.

$${Determine}\:{the}\:{complex}\:{number} \\ $$$${z}\:{such}\:{that}\:{z}^{\mathrm{2}} ={tanx}+{icosx}, \\ $$$${in}\:{the}\:{form}\:{z}={p}+{ik},\:{p},{k}\in{R}. \\ $$$$ \\ $$

Answered by prakash jain last updated on 26/Jan/15

(p+ik)^2 =tan x+i cos x (p^2 −k^2 )+2ipk=tan x+i cos x tan x=p^2 −k^2 ⇒k^2 =p^2 −tan x cos x=2pk cos^2 x=4p^2 (p^2 −tan x) 4p^4 −4p^2 tan x−cos^2 x=0 p^2 =((4tan x±(√(16tan^2 x+16cos^2 x)))/8) p^2 =((tan x+(√(tan^2 x+cos^2 x)))/2) k^2 =p^2 −tan x k^2 =(((√(tan^2 x+cos^2 x))−tan x)/2) p=±(√(((√(tan^2 x+cos^2 +))tan x)/2)) q=±(√(((√(tan^2 x+cos^2 x))−tan x)/2)) Check p^2 −q^2 =tan x 2pq=2×±(√(((√(tan^2 x+cos^2 x))+tan x)/2))×±(√(((√(tan^2 x+cos^2 x))+tan x)/2)) =(±×±)∙2∙(√((tan^2 x+ cos^2 x−tan^2 x)/4))=cos x

$$\left({p}+{ik}\right)^{\mathrm{2}} =\mathrm{tan}\:{x}+{i}\:\mathrm{cos}\:{x} \\ $$$$\left({p}^{\mathrm{2}} −{k}^{\mathrm{2}} \right)+\mathrm{2}{ipk}=\mathrm{tan}\:{x}+{i}\:\mathrm{cos}\:{x} \\ $$$$\mathrm{tan}\:{x}={p}^{\mathrm{2}} −{k}^{\mathrm{2}} \Rightarrow{k}^{\mathrm{2}} ={p}^{\mathrm{2}} −\mathrm{tan}\:{x} \\ $$$$\mathrm{cos}\:{x}=\mathrm{2}{pk} \\ $$$$\mathrm{cos}^{\mathrm{2}} {x}=\mathrm{4}{p}^{\mathrm{2}} \left({p}^{\mathrm{2}} −\mathrm{tan}\:{x}\right) \\ $$$$\mathrm{4}{p}^{\mathrm{4}} −\mathrm{4}{p}^{\mathrm{2}} \mathrm{tan}\:{x}−\mathrm{cos}^{\mathrm{2}} {x}=\mathrm{0} \\ $$$${p}^{\mathrm{2}} =\frac{\mathrm{4tan}\:{x}\pm\sqrt{\mathrm{16tan}^{\mathrm{2}} {x}+\mathrm{16cos}^{\mathrm{2}} {x}}}{\mathrm{8}} \\ $$$${p}^{\mathrm{2}} =\frac{\mathrm{tan}\:{x}+\sqrt{\mathrm{tan}^{\mathrm{2}} {x}+\mathrm{cos}^{\mathrm{2}} {x}}}{\mathrm{2}} \\ $$$${k}^{\mathrm{2}} ={p}^{\mathrm{2}} −\mathrm{tan}\:{x} \\ $$$${k}^{\mathrm{2}} =\frac{\sqrt{\mathrm{tan}^{\mathrm{2}} {x}+\mathrm{cos}^{\mathrm{2}} {x}}−\mathrm{tan}\:{x}}{\mathrm{2}} \\ $$$${p}=\pm\sqrt{\frac{\sqrt{\mathrm{tan}^{\mathrm{2}} {x}+\mathrm{cos}^{\mathrm{2}} +}\mathrm{tan}\:{x}}{\mathrm{2}}} \\ $$$${q}=\pm\sqrt{\frac{\sqrt{\mathrm{tan}^{\mathrm{2}} {x}+\mathrm{cos}^{\mathrm{2}} {x}}−\mathrm{tan}\:{x}}{\mathrm{2}}} \\ $$$$\mathrm{Check} \\ $$$${p}^{\mathrm{2}} −{q}^{\mathrm{2}} =\mathrm{tan}\:{x} \\ $$$$\mathrm{2}{pq}=\mathrm{2}×\pm\sqrt{\frac{\sqrt{\mathrm{tan}^{\mathrm{2}} {x}+\mathrm{cos}^{\mathrm{2}} {x}}+\mathrm{tan}\:{x}}{\mathrm{2}}}×\pm\sqrt{\frac{\sqrt{\mathrm{tan}^{\mathrm{2}} {x}+\mathrm{cos}^{\mathrm{2}} {x}}+\mathrm{tan}\:{x}}{\mathrm{2}}} \\ $$$$=\left(\pm×\pm\right)\centerdot\mathrm{2}\centerdot\sqrt{\frac{\mathrm{tan}^{\mathrm{2}} {x}+\:\mathrm{cos}^{\mathrm{2}} {x}−\mathrm{tan}^{\mathrm{2}} {x}}{\mathrm{4}}}=\mathrm{cos}\:{x} \\ $$

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