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So I'm trying to convert a complex number to the form of $a + bi$. The complex number in question is $$\bbox[5px,border:2px solid #C0A000]{\large 2e^{\frac{-3\pi}{4i}+\ln(3)}}$$ I'm not quite sure how to tackle this to be honest, I would appreciate some help in trying to understand how to convert these expressions into that form.

Thanks in advance!

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3 Answers 3

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there are two standard forms for complex numbers.

Rectangular form $z =a+bi$ where $a = Re(z)$ and $b = Im(z)$.

And, if $z \ne 0$ polar form $z = r e^{\theta i}$ where $r = |z|$ and $\theta =\arg(z)$.

$e^{\theta i}$ is defined to be $e^{\theta i} = \cos \theta + i \sin \theta$.

So there is a conversion between these two forms.

If $z = r e^{\theta i} = r\cos \theta + (r \sin \theta) i$.

And if $z = a + bi= \sqrt{a^2 + b^2} (\cos (\arctan \frac ba) + \sin (\arctan \frac ba)i) = \sqrt{a^2 + b^2} e^{\arctan \frac ba i}$.

....

Now the only trick this problem has done is it has combined then $r=|z|$ and $e^{\theta i}$ by replacing $r$ with $e^{\ln r}$ and combining $r e^{\theta i} = e^{\ln r} e^{\theta i} = e^{\theta i + \ln r}$ to get it as a single exponent.

.....

So $2e^{-3\pi/4i+ln(3)}=$

$2e^{-\frac {3\pi}4i}e^{\ln 3} =$

$6e^{-\frac {3pi}4i} = 6(\cos(-\frac {3\pi}4)+ i\sin (-\frac {3\pi} 4))=$

$6(-\frac {\sqrt 2}2 + \frac {\sqrt 2}2i) =$

$-3\sqrt 2 +3\sqrt 2 i$.

That's it.

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    $\begingroup$ am I mistaken or is your last conclusion incorrect? I think it should be $-3\sqrt{2} + 3\sqrt{2}i$? $\endgroup$
    – JakeDrone
    Jan 24, 2021 at 22:02
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    $\begingroup$ I make arithmetic mistakes all the time. But I do believe $\cos|\sin (-\frac {3\pi}4) =\pm\frac 1{\sqrt 2} =\pm \frac {\sqrt 2}2$. So if you multiply be $6$ you get $\pm 3\sqrt 2$. And I did type $\pm 3{\sqrt 2}$, didn't ? So I don't see how you'd get the $\pm \frac {3\sqrt 2}2$ that you typed. That is what I typed and what you typed aren't they?...... rereads what I typed and what you typed ..... GUDDAMMET!!!!!!... yes, you are right I made a typo. One should type $\frac 62 = 3$ with a $3$, not with $6$ (which I meant to be a $3$.... I swear I did.) $\endgroup$
    – fleablood
    Jan 24, 2021 at 22:23
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    $\begingroup$ Haha, good to know the more mathematically advanced make mistakes too :P thank you for your answer btw. $\endgroup$
    – JakeDrone
    Jan 24, 2021 at 22:29
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The first thing you can do is factor the expression into $2e^{\ln(3)} \times e^{-3\pi/4i}$.

Then, we can use Euler's formula: $e^{ix} = \cos x + i \sin x$.

If you apply that formula to the expression, you will be able to get your answer with a little bit of arithmetic and simplification.

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    $\begingroup$ so $e^{ix}$ is a formula that is known for these types of problems? $\endgroup$
    – JakeDrone
    Jan 24, 2021 at 21:35
  • $\begingroup$ Yup! That formula works for any real number $x$. $\endgroup$
    – inavda
    Jan 24, 2021 at 21:42
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    $\begingroup$ "so $ e^{ix}$ is a formula that is known for these types of problems?" VERY well known. Yes, you are expected to know this. You con always convert $e^{\theta i} = \cos \theta + i \sin \theta$ by definitions and so $e^{k + \theta i} = e^ke^{\theta i} = e^k\cos \theta + i e^k\sin \theta$..... And you can go the other way too: $a + bi = \sqrt{a^2 + b^2}(\frac a{\sqrt{a^2 + b^2}} + \frac b{\sqrt{a^2 + b^2}}i) = \sqrt{a^2 + b^2}(\cos \arctan \frac ba + i\sin \arctan \frac ba) = \sqrt{a^2 + b^2} e^{\arctan \frac ba i} = e^{\ln(\sqrt{a^2 +b^2}+ \arctan \frac ba i}$. $\endgroup$
    – fleablood
    Jan 24, 2021 at 21:49
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    $\begingroup$ @JakeDrone Actually, $$e^{ix} = \cos x + i \sin x$$ works for any $x$, which lets us extend $\cos$ & $\sin$ to the complex numbers. These functions are well-behaved over the entire complex plane. See en.wikipedia.org/wiki/Entire_function $\endgroup$
    – PM 2Ring
    Jan 24, 2021 at 22:46
  • $\begingroup$ Good catch -- right! $\endgroup$
    – inavda
    Jan 24, 2021 at 22:48
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First, convert it into $re^{i \theta}$. Then we have $a = r \cos \theta$ and $b = r \sin \theta$. In your case, we have $r = 2e^{\ln 3} = 6$ and $\theta = -3 \pi / 4$.

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