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Question Number 73697 by mathmax by abdo last updated on 14/Nov/19

$$findaformulaeforcalculusofarctan(x+iy)$$

Commented by mathmax by abdo last updated on 15/Nov/19

$$wehaveprovedthatarctanz=\frac{1}{2i}ln\left(\frac{1+iz}{1-iz}\right)\Rightarrow$$$$ln(x+iy)=\frac{1}{2i}ln\left(\frac{1+i(x+iy)}{1-i(x+iy)}\right)=\frac{1}{2i}ln\left(\frac{1-y+ix}{1+y-ix}\right)wehave$$$$\mid 1-y+ix\mid =\sqrt{{(1-y)}^2+x^2}\Rightarrow 1-y+ix=\sqrt{{(1-y)}^2+x^2}{e}^{iarctan\left(\frac{x}{1-y}\right)}$$$$\mid 1+y-ix\mid =\sqrt{{(1+y)}^2+x^2}\Rightarrow 1+y-ix=\sqrt{{(1+y)}^2+x^2}{e}^{-iarctan\left(\frac{x}{1+y}\right)}$$$$\Rightarrow arctan(x+iy)=\frac{1}{2i}ln\left(\frac{\sqrt{x^2+{(1-y)}^2}{e}^{iarctan\left(\frac{x}{1-y}\right)}}{\sqrt{x^2+{(y+1)}^2}{e}^{-iarctan\left(\frac{x}{1+y}\right)}}\right)$$$$=\frac{1}{4i}ln(x^2+{(1-y)}^2)+\frac{1}{2}arctan\left(\frac{x}{1-y}\right)-\frac{1}{4i}ln(x^2+{(y+1)}^2)$$$$+\frac{1}{2}arctan\left(\frac{x}{1+y}\right)$$$$=-\frac{i}{4}ln(x^2+{(1-y)}^2)+\frac{i}{4}ln(x^2+{(y+1)}^2)+\frac{1}{2}arctan\left(\frac{x}{1-y}\right)$$$$+\frac{1}{2}arctan\left(\frac{x}{1+y}\right)$$$$=\frac{1}{2}arctan\left(\frac{x}{1+y}\right)+\frac{1}{2}arctan\left(\frac{x}{1-y)}\right)+\frac{i}{4}ln\left(\frac{x^2+{(y+1)}^2}{x^2+{(1-y)}^2}\right)$$$$$$

Answered by Smail last updated on 15/Nov/19

$$arctan\left(x\right)=\frac{1}{2i}ln\left(\frac{x-i}{x+i}\right)$$$$soarctan(x+iy)=\frac{1}{2i}ln\left(\frac{x+iy-i}{x+iy+i}\right)$$$$ln\left(\frac{x+i(y-1)}{x+i(y+1)}\right)=a+ib$$$$\frac{x+i(y-1)}{x+i(y+1)}=e^a{e}^{ib}$$$$\frac{(x+i(y-1))(x+i(y+1))}{x^2+{(y+1)}^2}=e^a{e}^{ib}$$$$\frac{x^2+2ixy-(y^2-1)}{x^2+{(y+1)}^2}=e^a{e}^{ib}$$$$\frac{x^2-y^2+1}{x^2+{(y+1)}^2}+i\frac{2xy}{x^2+{(y+1)}^2}=e^a{e}^{ib}$$$$tan\left(b\right)=\frac{2xy}{x^2-y^2+1}\iff b=arctan\left(\frac{2xy}{x^2-y^2+1}\right)$$$$e^a=\sqrt{{\left(\frac{x^2-y^2+1}{x^2+{(y+1)}^2}\right)}^2+{\left(\frac{2xy}{x^2+{(y+1)}^2}\right)}^2}$$$$e^a=\frac{\sqrt{x^4+y^4-2x^2{y}^2+2x^2-2y^2+1+4x^2{y}^2}}{x^2+{(y+1)}^2}$$$$a=ln\left(\frac{\sqrt{{(x^2+y^2+1)}^2-4y^2}}{x^2+{(y+1)}^2}\right)$$$$tan(x+iy)=\frac{1}{2i}(ln\left(\frac{\sqrt{{(x^2+y^2+1)}^2-4y^2}}{x^2+{(y+1)}^2}\right)+iarctan\left(\frac{2xy}{x^2-y^2+1}\right)$$$$=\frac{1}{2}arctan\left(\frac{2xy}{x^2-y^2+1}\right)-i\frac{1}{2}ln\left(\frac{\sqrt{{(x^2+y^2+1)}^2-4y^2}}{x^2+{(y+1)}^2}\right)$$

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