#Complex Numbers Question
pale isle
I did this question assuming that $z^{-1}$ means the reciprocal of the complex number, such that it takes the form $\frac{1}{a+bi}$
torpid spindleBOT
martin(ping-me-when-reply)
pale isle
So if that is my mistake please let me know, I will now post the rest of my working
hearty timber
U are right haha
pale isle
:D
So I first wrote the number $(a+bi)^{-1} \$ in the form $\left(\sqrt{a^2+b^2}(\cos{\theta} + i\sin{\theta})\right)^{-1} \$ I then applied exponent laws and De Moivres theorem $\frac{1}{\sqrt{a^2+b^2}}(\cos{\theta}-i\sin{\theta}) \$ and then I took the real part and imaginary part? I do not know if this is legal lol $\ \Re{z^{-1}} = \frac{\cos{\theta}}{\sqrt{a^2+b^2}} \$ $\Im{z^{-1}} = -\frac{\sin{\theta}}{\sqrt{a^2+b^2}}
torpid spindleBOT
martin(ping-me-when-reply)
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pale isle
If this is correct, would there be any way to express the imaginary and real part of the number without theta?
PS- this textbook does not have any answers, and it seems to be a very random textbook that no one has posted answers for online,
crisp marlin
So I first wrote the number $(a+bi)^{-1} \\$ in the form $\left(\sqrt{a^2+b^2}(\...
...why don't you just do 1/(a + bi) * (a - bi)/(a - bi)?
hearty timber
**martin(ping-me-when-reply)**
Compile Error! Click the <:errors:504266556054962...
That's correct
hearty timber
If this is correct, would there be any way to express the imaginary and real par...
Yh, just do what @crisp marlin said
pale isle
$\frac{a-bi}{a^2+b^2} \$
$\frac{a}{a^2+b^2} = \Re{z^{-1}}$
$\frac{-b}{a^2+b^2} = \Im{z^{i1}}$
torpid spindleBOT
martin(ping-me-when-reply)
crisp marlin
Yeah.
hearty timber
Yep
pale isle
I was too caught up in what was covered in this chapter that I forgot to use the gooey brain matter I had the entire time
Thanks everyone
hearty timber
But the other method was also correct
crisp marlin
It's a poorly written textbook if the chapter on Euler's formula has a question about complex conjugates in it.
crisp marlin
But the other method was also correct
I don't think it was, though.
hearty timber
I don't think it was, though.
It was
crisp marlin
Oh, I see.
hearty timber
R*exp(it) goes to exp(-it)/R
crisp marlin
Yeah, it's the trig bits that threw me off, but now I see.
hearty timber
"Really basic complex stuff" lol
pale isle
It's a poorly written textbook if the chapter on Euler's formula has a question ...
The chapter was actually just about basic complex number stuff- but I had finished the part about complex congugates a while ago and only picked up the book today, so my brain was full of De Moivre's theorem when I started the questions and for some reason I tried to jam it in
I assume it's really basic idk it's the first time I have done anything with complex numbers
hearty timber
It's not basic
hearty timber
Cool
crisp marlin
If $a + bi = \sqrt{a^2 + b^2} * (\cos{\theta} + i\sin{\theta})$, then $\cos{\theta} + i\sin{\theta} = \frac{a}{\sqrt{a^2 + b^2}} + \frac{b}{\sqrt{a^2 + b^2}}i$.
torpid spindleBOT
Techie Literate
crisp marlin
Still feels like bringing an elephant gun to kill a fly.
hearty timber
Yh, it was basically that
The only good proof of why exp(it) is a good definition comes from differential equations
And that's not basic
crisp marlin
I mean. Differential calculus. Not differential equations.
crisp marlin
There's one I like that's even simpler.
crisp marlin
But the differential equations argument is the best
What's that, then?
crisp marlin
Okay, then.
hearty timber
Since cos(t) + i sin(t) satisfies the equation
We can define e^it by it's main property
crisp marlin
The one I like is this; if e^(ix) = cos(x) + isin(x), then e^(-ix) * (cos(x) + isin(x)) = 1. It's trivial to show that it equals 1 for x = 0, then you show the derivative is 0, so the function is a constant.