assuming $Z$ is a complex number i.e $a + bi$. we have

$Z^2 = (a + bi)^2 = (a^2-b^2)+(2ab)i$

$Z^3 = (a+bi)^3 = (a^3-3ab^2) - (b^3-3a^2b)i$

but what would be the general formula for computing $Z^n$ where n is any real number?

I wanted to use this method but i dont know how the computation goes for complex number $i$. also I want to do this computation on a computer so i have to keep real and imaginary parts separated from each other.

I know that i can only approximate numbers when $n$ is not an integer. but thats ok.

Is this even possible?

  • $\begingroup$ You can get these by using the binomial theorem. $\endgroup$ – Lord Shark the Unknown Apr 20 '18 at 17:47
  • 3
    $\begingroup$ Use the polar form of the complex numbers $\endgroup$ – Andrei Apr 20 '18 at 17:48
  • $\begingroup$ sorry the link was broken. fixed it. $\endgroup$ – M.kazem Akhgary Apr 20 '18 at 17:48
  • $\begingroup$ @Andrei good idea. i totally forgot that, thanks. $\endgroup$ – M.kazem Akhgary Apr 20 '18 at 17:51

If, when you say "I want to use this method", you mean that you really want to use the binomial expansion, rather than using polar form, then $(a+ ib)^n$ will be a sum of monomials of the form $\frac{a^(n-i)(ib)^i}{i!}$. "i" has "period 4". That is $i^1= i$, $i^2= -1$, $i^3= -1$, and $i^4= 1$ so that further powers are just repeats of that. You can separate that into one real part which will have powers that are multiples of 4 another subtracted with powers that are two more than multiples of 4, an imaginary part of two parts, the first with powers that are 1 more than multiples of 4, the second, subtracted with powers that are three more than powers of 4 : $\left(a^n+ \frac{a^{n-4}b^4}{4!}+ \frac{a^{n-8}b^8}{8!}+ \cdot\cdot\cdot+\frac{a^{n-4k}b^{4k}}{4k!}+ \cdot\cdot\cdot\right)- \left(a^{n-2}b^2+ \frac{a^{n- 6}b^6}{6!}+ \frac{a^{n-10}b^{10}}{10!}+ \cdot\cdot\cdot+\frac{a^{n-4k- 2}b^{4k+2}}{(4k+2)!}+ \cdot\cdot\cdot\right)+ i\left(a^{n-1}+ \frac{a^{n-5}b^5}{5!}+ \frac{a^{n-9}b^9}{9!}+ \cdot\cdot\cdot+\frac{a^{n-4k-1}b^{4k+1}}{(4k+1)!}+ \cdot\cdot\cdot\right)- i\left(a^n+ \frac{a^{n-4}b^4}{4!}+ \frac{a^{n-8}b^8}{8!}+ \cdot\cdot\cdot+\frac{a^{n-4k-3}b^{4k+3}}{(4k+3)!}+ \cdot\cdot\cdot\right)$.

  • $\begingroup$ in my program i had to use $atan2$ instead of $atan$ but that totally worked. $\endgroup$ – M.kazem Akhgary Apr 20 '18 at 18:45
  • $\begingroup$ your previous answer was the one that helped me. please keep that (as well as this if you want.) thanks for the effort. and sorry i really didnt know the main solution. polar form that you provided was much simpler to achieve. $\endgroup$ – M.kazem Akhgary Apr 20 '18 at 18:48
  • $\begingroup$ enjoy :) you can set the power to a constant value as well. glslsandbox.com/e#46501.8 $\endgroup$ – M.kazem Akhgary Apr 20 '18 at 18:58

For $z=r(\cos\phi+i\sin\phi)$, where $r=\sqrt{a^2+b^2}$, $\cos\phi=\frac{a}{r}$, $\sin\phi=\frac{b}{r}$ we have $$ z^n=r(\cos n\phi+i\sin n\phi) $$ It is a basic method.

  • $\begingroup$ never thought it would be this easy. thanks a lot $\endgroup$ – M.kazem Akhgary Apr 20 '18 at 17:53

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