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I need to convert the following complex number to the algebraic form but I need help.

Should I use the trigonometric form firstly to find the module and argument or is there another method

$$z=\frac{e^{1+2i}-1}{e^{1+2i}+1}$$

Thank you for your help!

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    $\begingroup$ Multiply numerator and denominator by the conjugate of the denominator. $\endgroup$
    – egreg
    Apr 26, 2017 at 22:41

2 Answers 2

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$$z=\frac{e^{1+2i}-1}{e^{1+2i}+1}=\frac{e^{1+2i}+1-2}{e^{1+2i}+1}=1-\frac{2}{e^{1+2i}+1}=1-\frac{2}{e \cdot e^{2i}+1}$$

$$=1-\frac{2}{e \cdot (\cos 2 + i \sin 2)+1}=1-\frac{2}{(e\cos 2+1) + i e \sin 2}$$

Now multiply and divide by the conjugate of $$(e\cos 2+1) + i e \sin 2$$

This is probably the shortest way I could solve.

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The conjugate of $e^{1+2i}+1$ is $e^{1-2i}+1$. Thus $$ \frac{e^{1+2i}-1}{e^{1+2i}+1}= \frac{e^{1+2i}-1}{e^{1+2i}+1}\frac{e^{1-2i}+1}{e^{1-2i}+1}= \frac{e^2+e^{1+2i}-e^{1-2i}-1}{e^2+e^{1+2i}+e^{1-2i}+1} $$ Now note that $$ e^{1+2i}-e^{1-2i}=2ie\sin2 \qquad e^{1+2i}+e^{1-2i}=2e\cos2 $$

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