Question Number 186960 by Humble last updated on 12/Feb/23
Answered by MJS_new last updated on 12/Feb/23
![z=x+yi z=∣z∣e^(∡(z)) [∣x+yi∣=(√(x^2 +y^2 )) ∧ ∡(x+yi)=arctan (y/x)] z=(√(x^2 +y^2 ))e^(i arctan (y/x)) ln z =ln ((√(x^2 +y^2 ))e^(i arctan (y/x)) ) [ln ab =ln a +ln b] ln z =ln (√(x^2 +y^2 )) +ln e^(i arctan (y/x)) [ln (√a) =(1/2)ln a ∧ ln e^b =b] ln z =(1/2)ln (x^2 +y^2 ) +i arctan (y/x)](https://www.tinkutara.com/question/Q187002.png)
$${z}={x}+{y}\mathrm{i} \\ $$$${z}=\mid{z}\mid\mathrm{e}^{\measuredangle\left({z}\right)} \\ $$$$\:\:\:\:\:\left[\mid{x}+{y}\mathrm{i}\mid=\sqrt{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }\:\wedge\:\measuredangle\left({x}+{y}\mathrm{i}\right)=\mathrm{arctan}\:\frac{{y}}{{x}}\right] \\ $$$${z}=\sqrt{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }\mathrm{e}^{\mathrm{i}\:\mathrm{arctan}\:\frac{{y}}{{x}}} \\ $$$$\mathrm{ln}\:{z}\:=\mathrm{ln}\:\left(\sqrt{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }\mathrm{e}^{\mathrm{i}\:\mathrm{arctan}\:\frac{{y}}{{x}}} \right) \\ $$$$\:\:\:\:\:\left[\mathrm{ln}\:{ab}\:=\mathrm{ln}\:{a}\:+\mathrm{ln}\:{b}\right] \\ $$$$\mathrm{ln}\:{z}\:=\mathrm{ln}\:\sqrt{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }\:+\mathrm{ln}\:\mathrm{e}^{\mathrm{i}\:\mathrm{arctan}\:\frac{{y}}{{x}}} \\ $$$$\:\:\:\:\:\left[\mathrm{ln}\:\sqrt{{a}}\:=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\:{a}\:\wedge\:\mathrm{ln}\:\mathrm{e}^{{b}} \:={b}\right] \\ $$$$\mathrm{ln}\:{z}\:=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\:\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)\:+\mathrm{i}\:\mathrm{arctan}\:\frac{{y}}{{x}} \\ $$
Commented by Humble last updated on 12/Feb/23
$${thank}\:{you}\:{sir} \\ $$