I would've rewritten the left-hand side as z^a*(z^bi). Then z^bi as cosb + i*sinb so z^a(cosb + i*sinb) Then z^a (cosb)+ z^a (sinb)= c+di z^a(cosb)=c and z^a(sinb)=d. Finding b should be a matter of setting sinb/cosb = d/c. I'm not sure how to go about finding a from here, though.😔
4:43 θ=arctan(d/c)+n2π ? But is it missing in the 9:19 solution? I hoped to see all of the solutions. Maybe if its should be arg(c,d) rather than arctan(d/c) if the arg(c,d) or arg(c+di) is considered as valid mathematical function....
Wolfram Alpha's result 😂🤣 It is like all the math teacher humans wanting an exact formula and a calculator or computing device resistance to absurdity in human demands giving back to the human only the real answers of doing an approximation in decimal numbers (at the most incorporating binary equivalent to human base 10 counting systems) using calculators or computer devices with no stunning exact formula insights. 😂🤣👍
This was interesting since seeing your infinite power Z = i problem formed Z^i = i that doesn't seem as a strange I think I knew what is wrong but confused in my mind without other proof. Now I'm trying to digest your formula that doesn't match what you found days ago in an infinite tower Z = i Tower Power. Your derived a middle solution to today's formula now with a=0, b=1, c=0, d=1 and your Z solution becomes: Z = (0 + π/2 + i(0 + 1(0)) !? Z = π/2 is not e^π/2 ?? I think that in today's formula we have an e^Z solution you came up with in the infinite power problem days ago. 🤔😑
Messy, but interesting!
now try to come up with a formula for
(a+bi)^(c+di)=x+yi
solve for x and y which are real numbers
I would've rewritten the left-hand side as z^a*(z^bi). Then z^bi as cosb + i*sinb so z^a(cosb + i*sinb)
Then z^a (cosb)+ z^a (sinb)= c+di
z^a(cosb)=c and z^a(sinb)=d. Finding b should be a matter of setting sinb/cosb = d/c. I'm not sure how to go about finding a from here, though.😔
your second step only works if z=e.
@@decare696 You're right. I didn't catch that until hours later. Z^(bi) should be e^(bi*ln(z)).
4:43 θ=arctan(d/c)+n2π ?
But is it missing in the 9:19 solution?
I hoped to see all of the solutions.
Maybe if its should be arg(c,d) rather than arctan(d/c) if the arg(c,d) or arg(c+di) is considered as valid mathematical function....
There's no way to clean that dirty fuzzball.
Wolfram alpha development jokes upon their users 😅
Wolfram Alpha's result 😂🤣 It is like all the math teacher humans wanting an exact formula and a calculator or computing device resistance to absurdity in human demands giving back to the human only the real answers of doing an approximation in decimal numbers (at the most incorporating binary equivalent to human base 10 counting systems) using calculators or computer devices with no stunning exact formula insights. 😂🤣👍
Why don't you like WolframAlpha's solution? After all, it is correct.
This was interesting since seeing your infinite power Z = i problem formed Z^i = i that doesn't seem as a strange I think I knew what is wrong but confused in my mind without other proof. Now I'm trying to digest your formula that doesn't match what you found days ago in an infinite tower Z = i Tower Power.
Your derived a middle solution to today's formula now with a=0, b=1, c=0, d=1 and your Z solution becomes:
Z = (0 + π/2 + i(0 + 1(0)) !?
Z = π/2 is not e^π/2 ??
I think that in today's formula we have an e^Z solution you came up with in the infinite power problem days ago. 🤔😑
My bad! 😬😳 I wrote down Z = formula vs seeing the video again and knowing you were solving for ln(Z) 😑
You did not express z in the form of x + yi
Did I not? 🤪
Looks unwieldy. Let's invent a third notation for complex numbers that makes exponentiation simple.