How to write $4-3i$ in the form $e^z$ with $z=x+yi$

Question

I need to write $4-3i$ in the form of $e^z$ with $z=x+yi$. Important is that it shouldn't be like $z=\ln(\cdot)$. The exercise demands this. The problem is that I do not know how to evade the $\ln()$ form.

I've only managed to do this: $|4-3i|= 5$. $\text{Arg}(4-3i) = \tan^{-1}(-3/4)$.

So $e^x=5 \hspace{10mm} e^y = \tan^{-1}(-3/4)$

$x=\ln(5) \hspace{10mm} y=\ln(-3/4)$

Anyone have any idea how to evade the $\ln(\cdot)$?

Answer 1

$$ e^{x+iy} = e^x ( \cos y + i \sin y ) $$ You are correct about the $5$ $$ 4 - 3 i = 5 ( \cos y + i \sin y ) $$ $$ \cos y = \frac{4}{5} \; , \; \; \; \sin y = -\frac{3}{5} $$ $$ \tan y = -\frac{3}{4} $$

Meanwhile, tangent is negative in either the second or fourth quadrant. Here we can use only the fourth quadrant. As $\arctan$ gives either fourth or first quadrant, we can use it, then add integer multiples of $2 \pi$ to $y$

Answer 2

You want to write it as $z=|z|e^{i\theta}$

$$|z|=\sqrt{4^2+3^2}=5$$ $$arg(z)=\theta=\tan^{-1}\left(\frac{-3}{4}\right)$$

thus you can write it as

$$z=5\cdot e^{i\cdot tan^{-1}\frac{-3}{4}}$$ $$z=e^{ln5}\cdot e^{i\cdot tan^{-1}\frac{-3}{4}}$$ $$z= 4-3i=e^{ln5+i\cdot tan^{-1}\frac{-3}{4}}$$

$$e^{x+iy}=e^{\ln5+i\cdot tan^{-1}\frac{-3}{4}}$$ $$x=\ln5$$ $$y=tan^{-1}\frac{-3}{4}$$

Answer 3

You need to find (a) $z=x+i y$ such that $$e^{z}=4-3i\implies e^{x+iy}=4-3i.$$ Now, from the Euler formula $e^{z}=e^{x+iy}=e^{x}e^{i y}=e^{x}(\cos(y)+i\sin(y))$ so the equation becomes $$e^{x}(\cos(y)+i\sin(y))=4-3i$$. Observe now that the modulus of RHS is equal to $e^{x}$, in fact: $$|e^{x}\cos(y)+ie^{x}\sin(y)|=\sqrt{e^{2x}\cos^2(y)+e^{2x}\sin^2(y)}=\sqrt{e^{2x}}=e^{x}\ \ \ \forall x\in\mathbb{R}.$$ The modulus of $4-3i$ is: $$|4-3i|=\sqrt{25}=5,$$ hence $$|e^{z}|=|4-3i|\iff e^{x}=5\iff x=\ln(5).$$

Now we have to find $y$ such that $e^{\ln(5)}(\cos(y)+i\sin(y))=4-3i$ which means that we have to solve $$\begin{cases}5\cos(y)=4\\ 5\sin(y)=-3\end{cases}\implies y=-\arctan\left(\frac{3}{4}\right)+2k\pi\ \ \ k\in\mathbb{Z}$$. So

$$z=\ln(5)+i\left(-\arctan\left(\frac{3}{4}\right)+2k\pi\right)\ \ \ \mbox{with} \ k\in\mathbb{Z}.$$