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If-z-cos-isin-0-lt-lt-pi-6-then-prove-that-argument-of-1-z-4-2-pi-2-

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Question Number 43363 by rahul 19 last updated on 10/Sep/18
If z= cos θ + isin θ , 0<θ<(π/6) , then prove that argument of 1−z^4 = 2θ − (π/2) .
$$\mathrm{If}\:\mathrm{z}=\:\mathrm{cos}\:\theta\:+\:\mathrm{isin}\:\theta\:,\:\mathrm{0}<\theta<\frac{\pi}{\mathrm{6}}\:,\:\mathrm{then}\:\mathrm{prove} \\ $$$$\mathrm{that}\:\mathrm{argument}\:\mathrm{of}\:\:\mathrm{1}−\mathrm{z}^{\mathrm{4}} \:=\:\mathrm{2}\theta\:−\:\frac{\pi}{\mathrm{2}}\:. \\ $$
Commented by maxmathsup by imad last updated on 10/Sep/18
we have z =e^(iθ) ⇒1−z^4 =1−e^(4iθ) =1−cos(4θ)−isin(4θ) =2 sin^2 (2θ) −2isin(2θ)cos(2θ) =−2isin(2θ){cos(2θ) +isin(2θ)} =2 sin(2θ)(−i) e^(i(2θ)) we have 0<2θ<(π/3) ⇒sin(2θ)>0 ⇒∣1−z^4 ∣ =2 sin(2θ) and arg(1−z^4 ) ≡arg(−i) +2θ [2π] ⇒arg(1−z^4 )≡ −(π/2) +2θ[2π] .
$${we}\:{have}\:{z}\:={e}^{{i}\theta} \:\Rightarrow\mathrm{1}−{z}^{\mathrm{4}} \:=\mathrm{1}−{e}^{\mathrm{4}{i}\theta} \:\:\:=\mathrm{1}−{cos}\left(\mathrm{4}\theta\right)−{isin}\left(\mathrm{4}\theta\right) \\ $$$$=\mathrm{2}\:{sin}^{\mathrm{2}} \left(\mathrm{2}\theta\right)\:−\mathrm{2}{isin}\left(\mathrm{2}\theta\right){cos}\left(\mathrm{2}\theta\right)\:=−\mathrm{2}{isin}\left(\mathrm{2}\theta\right)\left\{{cos}\left(\mathrm{2}\theta\right)\:+{isin}\left(\mathrm{2}\theta\right)\right\} \\ $$$$=\mathrm{2}\:{sin}\left(\mathrm{2}\theta\right)\left(−\boldsymbol{{i}}\right)\:\boldsymbol{{e}}^{\boldsymbol{{i}}\left(\mathrm{2}\theta\right)} \:\:{we}\:{have}\:\:\mathrm{0}<\mathrm{2}\theta<\frac{\pi}{\mathrm{3}}\:\Rightarrow{sin}\left(\mathrm{2}\theta\right)>\mathrm{0}\:\Rightarrow\mid\mathrm{1}−{z}^{\mathrm{4}} \mid\:=\mathrm{2}\:{sin}\left(\mathrm{2}\theta\right) \\ $$$${and}\:\:{arg}\left(\mathrm{1}−{z}^{\mathrm{4}} \right)\:\equiv{arg}\left(−{i}\right)\:+\mathrm{2}\theta\:\:\left[\mathrm{2}\pi\right]\:\Rightarrow{arg}\left(\mathrm{1}−{z}^{\mathrm{4}} \right)\equiv\:−\frac{\pi}{\mathrm{2}}\:+\mathrm{2}\theta\left[\mathrm{2}\pi\right]\:. \\ $$
Answered by ajfour last updated on 10/Sep/18
1−z^4 =1−(x+iy)^4 = (1−x^4 −y^4 +6x^2 y^2 )+4i(xy^3 −x^3 y) tan φ = ((4xy(y^2 −x^2 ))/(1−(x^2 −y^2 )^2 +4x^2 y^2 )) = ((4m(m^2 −1))/((1+m^2 )^2 −(1−m^2 )^2 +4m^2 )) (if m=tan θ ) ⇒ tan φ = ((m^2 −1)/(2m)) = −cot 2θ = tan (2θ−(π/2)) ⇒ φ (argument of 1−z^4 )= 2θ−(π/2) .
$$\mathrm{1}−{z}^{\mathrm{4}} =\mathrm{1}−\left({x}+{iy}\right)^{\mathrm{4}} \\ $$$$\:\:=\:\left(\mathrm{1}−{x}^{\mathrm{4}} −{y}^{\mathrm{4}} +\mathrm{6}{x}^{\mathrm{2}} {y}^{\mathrm{2}} \right)+\mathrm{4}{i}\left({xy}^{\mathrm{3}} −{x}^{\mathrm{3}} {y}\right) \\ $$$$\:\:\mathrm{tan}\:\phi\:=\:\frac{\mathrm{4}{xy}\left({y}^{\mathrm{2}} −{x}^{\mathrm{2}} \right)}{\mathrm{1}−\left({x}^{\mathrm{2}} −{y}^{\mathrm{2}} \right)^{\mathrm{2}} +\mathrm{4}{x}^{\mathrm{2}} {y}^{\mathrm{2}} } \\ $$$$\:\:\:\:=\:\frac{\mathrm{4}{m}\left({m}^{\mathrm{2}} −\mathrm{1}\right)}{\left(\mathrm{1}+{m}^{\mathrm{2}} \right)^{\mathrm{2}} −\left(\mathrm{1}−{m}^{\mathrm{2}} \right)^{\mathrm{2}} +\mathrm{4}{m}^{\mathrm{2}} } \\ $$$$\:\:\:\:\left({if}\:\:{m}=\mathrm{tan}\:\theta\:\right) \\ $$$$\Rightarrow\:\:\:\mathrm{tan}\:\phi\:=\:\frac{{m}^{\mathrm{2}} −\mathrm{1}}{\mathrm{2}{m}}\:=\:−\mathrm{cot}\:\mathrm{2}\theta \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\mathrm{tan}\:\left(\mathrm{2}\theta−\frac{\pi}{\mathrm{2}}\right) \\ $$$$\Rightarrow\:\phi\:\left({argument}\:{of}\:\mathrm{1}−{z}^{\mathrm{4}} \right)=\:\mathrm{2}\theta−\frac{\pi}{\mathrm{2}}\:. \\ $$
Commented by rahul 19 last updated on 10/Sep/18
thanks sir.
$$\mathrm{thanks}\:\mathrm{sir}. \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 10/Sep/18
z=cosθ+isinθ=e^(iθ) 1−z^4 =1−e^(i4θ) =1−(cos4θ+isin4θ) =1−cos4θ−isin4θ =2sin^2 2θ−i×2sin2θ.cos2θ so tanα=((−2sin2θcos2θ)/(2sin^2 2θ))=−cot2θ tan∝=−tan((Π/2)−2θ) tanα=tan(2θ−(Π/2)) α=2θ−(Π/2)
$${z}={cos}\theta+{isin}\theta={e}^{{i}\theta} \\ $$$$\mathrm{1}−{z}^{\mathrm{4}} \\ $$$$=\mathrm{1}−{e}^{{i}\mathrm{4}\theta} \\ $$$$=\mathrm{1}−\left({cos}\mathrm{4}\theta+{isin}\mathrm{4}\theta\right) \\ $$$$=\mathrm{1}−{cos}\mathrm{4}\theta−{isin}\mathrm{4}\theta \\ $$$$=\mathrm{2}{sin}^{\mathrm{2}} \mathrm{2}\theta−{i}×\mathrm{2}{sin}\mathrm{2}\theta.{cos}\mathrm{2}\theta \\ $$$${so}\:{tan}\alpha=\frac{−\mathrm{2}{sin}\mathrm{2}\theta{cos}\mathrm{2}\theta}{\mathrm{2}{sin}^{\mathrm{2}} \mathrm{2}\theta}=−{cot}\mathrm{2}\theta \\ $$$${tan}\propto=−{tan}\left(\frac{\Pi}{\mathrm{2}}−\mathrm{2}\theta\right) \\ $$$${tan}\alpha={tan}\left(\mathrm{2}\theta−\frac{\Pi}{\mathrm{2}}\right) \\ $$$$\alpha=\mathrm{2}\theta−\frac{\Pi}{\mathrm{2}} \\ $$
Commented by rahul 19 last updated on 10/Sep/18
thanks sir.
$$\mathrm{thanks}\:\mathrm{sir}. \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 10/Sep/18
z=cosθ+isinθ=e^(iθ) 1−z^4 =1−e^(i4θ) =1−(cos4θ+isin4θ) =1−cos4θ−isin4θ =2sin^2 2θ−i×2sin2θ.cos2θ so tanα=((−2sin2θcos2θ)/(2sin^2 2θ))=−cot2θ tan∝=−tan((Π/2)−2θ) tanα=tan(2θ−(Π/2)) α=2θ−(Π/2)
$${z}={cos}\theta+{isin}\theta={e}^{{i}\theta} \\ $$$$\mathrm{1}−{z}^{\mathrm{4}} \\ $$$$=\mathrm{1}−{e}^{{i}\mathrm{4}\theta} \\ $$$$=\mathrm{1}−\left({cos}\mathrm{4}\theta+{isin}\mathrm{4}\theta\right) \\ $$$$=\mathrm{1}−{cos}\mathrm{4}\theta−{isin}\mathrm{4}\theta \\ $$$$=\mathrm{2}{sin}^{\mathrm{2}} \mathrm{2}\theta−{i}×\mathrm{2}{sin}\mathrm{2}\theta.{cos}\mathrm{2}\theta \\ $$$${so}\:{tan}\alpha=\frac{−\mathrm{2}{sin}\mathrm{2}\theta{cos}\mathrm{2}\theta}{\mathrm{2}{sin}^{\mathrm{2}} \mathrm{2}\theta}=−{cot}\mathrm{2}\theta \\ $$$${tan}\propto=−{tan}\left(\frac{\Pi}{\mathrm{2}}−\mathrm{2}\theta\right) \\ $$$${tan}\alpha={tan}\left(\mathrm{2}\theta−\frac{\Pi}{\mathrm{2}}\right) \\ $$$$\alpha=\mathrm{2}\theta−\frac{\Pi}{\mathrm{2}} \\ $$

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