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Question Number 73113 by TawaTawa last updated on 06/Nov/19

Commented by mathmax by abdo last updated on 06/Nov/19

$$1)wehavearg\left(z\right)=arg(7-3\sqrt{3}i)+arg(-1-i)\left[2\pi \right]$$$$\mid 7-3\sqrt{3}\mid =\sqrt{49+27}=\sqrt{76}\Rightarrow 7-3\sqrt{3}=\sqrt{76}{e}^{iarctan\left(\frac{-3\sqrt{3}}{7}\right)}\Rightarrow$$$$arg(7-3\sqrt{3})=-arctan\left(\frac{3\sqrt{3}}{7}\right)$$$$-1-i=\sqrt{2}(-\frac{1}{\sqrt{2}}-\frac{i}{\sqrt{2}})=\sqrt{2}{e}^{i\frac{5\pi }{4}}\Rightarrow arg\left(z\right)=\frac{5\pi }{4}-arctan\left(\frac{3\sqrt{3}}{7}\right)\left[2\pi \right]$$$$$$

Commented by TawaTawa last updated on 06/Nov/19

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}$$

Commented by mathmax by abdo last updated on 06/Nov/19

$$youarewelcome.$$

Answered by mr W last updated on 06/Nov/19

$$\left(2\right)$$$$f\left(x\right)={(x-a)}^3+{(x-b)}^3+{(x-c)}^3$$$$\underset{x\to -\mathrm{\infty }}{\lim }f\left(x\right)=-\mathrm{\infty }$$$$\underset{x\to +\mathrm{\infty }}{\lim }f\left(x\right)=+\mathrm{\infty }$$$$f^{\prime }\left(x\right)=3{(x-a)}^2+3{(x-b)}^2+3{(x-c)}^2>0$$$$\Rightarrow f\left(x\right)isstrictlyincreasing,$$$$\Rightarrow f\left(x\right)=0hasoneandonlyonereal$$$$rootin(-\mathrm{\infty },+\mathrm{\infty }).$$

Commented by TawaTawa last updated on 06/Nov/19

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}$$

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