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Question Number 186960 by Humble last updated on 12/Feb/23

Answered by MJS_new last updated on 12/Feb/23

$$z=x+y\mathrm{i}$$$$z=\mid z\mid {\mathrm{e}}^{\measuredangle \left(z\right)}$$$$[\mid x+y\mathrm{i}\mid =\sqrt{x^2+y^2}\wedge \measuredangle (x+y\mathrm{i})=\arctan \frac{y}{x}]$$$$z=\sqrt{x^2+y^2}{\mathrm{e}}^{\mathrm{i}\arctan \frac{y}{x}}$$$$\ln z=\ln \left(\sqrt{x^2+y^2}{\mathrm{e}}^{\mathrm{i}\arctan \frac{y}{x}}\right)$$$$[\ln ab=\ln a+\ln b]$$$$\ln z=\ln \sqrt{x^2+y^2}+\ln {\mathrm{e}}^{\mathrm{i}\arctan \frac{y}{x}}$$$$[\ln \sqrt{a}=\frac{1}{2}\ln a\wedge \ln {\mathrm{e}}^b=b]$$$$\ln z=\frac{1}{2}\ln (x^2+y^2)+\mathrm{i}\arctan \frac{y}{x}$$

Commented by Humble last updated on 12/Feb/23

$$thankyousir$$

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