Solving $ e^z + 2 = 0$

Question

Why is this true?

If $\; e^z + 2 = 0$ and $z \in \Bbb C$, $\;$ then $\;$ $z = \ln 2 + i\,\pi$.

Probably simple question but I am not sure how to deal with it.

Answer 1

Because $$z=\ln 2 +i \pi \implies e^z =e^{\ln 2+i \pi}=e^{\ln 2}e^{i\pi}=2 \cdot(-1)=-2,$$ so $e^z+2=0$.

I used the identity $e^{i \pi}=-1$.

I just noticed that you wanted to imply the first from the latter, this is the other way around, since the other implication does not hold in general.

Answer 2

Hint Note that $e^z+2=0$ if and only if $e^z = -2 = -2+i\cdot 0 $ and use Euler's formla: $$e^{x+i y} = e^x(\cos(y)+i\sin(y))\qquad \forall x,y\in \Bbb R.$$

Answer 3

$$e^z+2=0\implies e^z=-2\,,\,k\in\Bbb Z\implies$$

$$z=\text{Log}(-2)=\log|-2|+i\arg(-2)=\log2+i\pi$$

When in the above we take the principal arugment.

Answer 4

Notice that if we have some: $z\in\mathbb{C}$ we can represent this as $z=x+jy$ where $x,y\in\mathbb{R}$. So if we want to take the natural logarithm we must instead use the polar system: $$\ln(z)=\ln(x+jy)=\ln(re^{j\theta})$$ And this can easily be split up into: $$\ln(re^{j\theta})=\ln(r)+j\theta$$ In this context we have: $$e^z+2=0\implies z=\ln(-2)$$ And we can say that: $$-2=2e^{j\pi}\therefore \ln(-2)=\ln(2)+j\pi$$ And so: $$z=\ln(2)+j\pi$$ However note that the imaginary part of this can make many forms since the function $e^z$ is periodic as can be seen by its definition $e^{j\theta}=\cos\theta+j\sin\theta$