5
$\begingroup$

Am I on the right track to solving this?

$$e^z=6i$$ Let $w=e^z$

Thus,

$$w=6i$$ $$e^w=e^{6i}$$ $$e^w=\cos(6)+i\sin(6)$$ $$\ln(e^w)=\ln(\cos(6)+i\sin(6))$$ $$w=\ln(\cos(6)+i\sin(6))$$ $$e^z=\ln(\cos(6)+i\sin(6))$$ $$\ln(e^z)=\ln(\ln(\cos(6)+i\sin(6)))$$ $$z=\ln(\ln(\cos(6)+i\sin(6)))$$

I had another method that started by taking the natural log of both sides right away, but that leads to $\arctan(6/0)$, which is undefined...

$\endgroup$
  • 1
    $\begingroup$ Write $6i$ in the form $e^{ia}$ for a better solution $\endgroup$ – George Feb 13 '16 at 0:20
  • $\begingroup$ $i=e^{\frac{i\pi}{2}}$ $\endgroup$ – Eleven-Eleven Feb 13 '16 at 0:21
  • $\begingroup$ @George isn't that what I did for step #2 after the word "Thus," ? $\endgroup$ – whatwhatwhat Feb 13 '16 at 0:23
  • 1
    $\begingroup$ @whatwhatwhat No, you made both sides to the equation an exponent of $e$. George wants you to write $6i=e^a e^{bi}=e^{a+bi}$. Here, $e^a$ is the real coefficient, so $e^a=6$ and $e^{bi}$ is the imaginary coefficient, so $e^{bi}=i$. As George said, $a=\ln 6$ and $b=\frac \pi 2$. $\endgroup$ – Noble Mushtak Feb 13 '16 at 0:26
  • 2
    $\begingroup$ What @Eleven-Eleven refers to is related to Euler's identity, which is very useful for solving this problem. It's what you used to change $e^{6i}=\cos(6)+i\sin(6)$. $\endgroup$ – Noble Mushtak Feb 13 '16 at 0:36
5
$\begingroup$

$e^z=6i$.

Let $z=x+iy$. Note that $e^z=e^x\cdot e^{iy}$

Thus $$e^z=e^x\cdot e^{iy}=6e^{i\left(\frac{\pi}{2}+2k\pi\right)}$$

So $e^x=6$ and so $x=\ln{6}$.

So $y=\frac{\pi}{2}+2k\pi$

Therefore you have as your solutions $z=\ln{6}+i\left(\frac{\pi}{2}+2k\pi\right)$ for integer $k$.

$\endgroup$
  • $\begingroup$ Why only $\frac \pi 2+2\pi$? Why not just $\frac \pi 2$ or list all solutions with $\frac \pi 2+2\pi k$? $\endgroup$ – Noble Mushtak Feb 13 '16 at 0:28
  • $\begingroup$ needed the $k$, thanks. I just missed it typing fast. $\endgroup$ – Eleven-Eleven Feb 13 '16 at 0:29
  • $\begingroup$ Wait, this is great and I thank you, but what was wrong with my solution? $\endgroup$ – whatwhatwhat Feb 13 '16 at 0:32
  • $\begingroup$ I get that it technically doesn't feel like it's fulfilling the "Find all $z$ such that..." $\endgroup$ – whatwhatwhat Feb 13 '16 at 0:32
  • 1
    $\begingroup$ The $k$ helps represent the infinite solutions for $z$. $\endgroup$ – Eleven-Eleven Feb 13 '16 at 0:33
4
$\begingroup$

Hopefully, from all of these solutions, you know how to solve this problem. Now, let's try doing it your way. You've done everything right so far: $$z=\ln(\ln(\cos(6)+i\sin(6)))$$

By Euler's Identity, we have $\cos(6)+i\sin(6)=e^{6i}$, so clearly, taking the $\ln$ of this is just $6i$: $$z=\ln(6i)$$

Now, if we go back to our original equation: $$e^z=6i$$

The equation we have at the end of all of this is just taking the $\ln$ of both sides of the original equation. Basically, everything you did is all valid, but you basically return to the original equation when we're all done with simplifying everything, which is why you were off-track.

$\endgroup$
  • 1
    $\begingroup$ Ahh...that makes a ton of sense. Thanks! $\endgroup$ – whatwhatwhat Feb 13 '16 at 0:43
2
$\begingroup$

Suppose $z=x+iy$ and $x$ and $y$ are real. Then $$ 6i = 6(0 + i) = e^z = e^{x+iy} = e^x e^{iy} = e^x(\cos y + i\sin y). $$ So $e^x = 6$ and $0+1i=\cos y + i\sin y$. Thus $\cos y=0$ and $\sin y=1$. So $y = \pi/2$ or $\pi/2+ 2\pi n$ for some integer $n$.

$\endgroup$
2
$\begingroup$

HINT:

$$6i=e^{\log(6)+i\pi/2+i2n\pi}=e^z$$

$\endgroup$
  • 1
    $\begingroup$ Ah!, first time I've seen you give a solution which may be a little difficult for the OP to understand !. Anyways, +1 $\endgroup$ – Shailesh Feb 13 '16 at 0:40
  • $\begingroup$ Shouldn't the last bit in the exponents be just $2\pi n$? Why is there an $i$ there? $\endgroup$ – whatwhatwhat Feb 13 '16 at 0:49
  • $\begingroup$ @shailesh Thank you for the up vote! Much appreciative. - Mark $\endgroup$ – Mark Viola Feb 13 '16 at 0:50
  • $\begingroup$ @whatwhatwhat The $i$ is correct since $e^{i2n\pi}=1$ for all integer values of $n$. But $e^{2\pi n}>1$ for $n\ne0$. $\endgroup$ – Mark Viola Feb 13 '16 at 0:52
  • $\begingroup$ Ok so you didn't just use the identity $\ln(z)=\ln|z|+iarg(z)+2\pi k$ then? $\endgroup$ – whatwhatwhat Feb 13 '16 at 0:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.