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My question is about raising a complex number to a high power, I know how to do that with De Moivre law, but i need to get the result in cartesian form, like $z=x+iy$. and without trigonometric terms.

The problem exactly is:

Write the following complex number in the following form $z=x+iy$: $$(3-2i)^3\cdot(1-i)^9$$

I searched alot the web to find an understandible explanation to solve the problem, but still i have no idea how to do that. Is there a special formula to raise a complex number to high power?

Thanks for help!!

Nosrati
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D.Rotnemer
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  • The best way is by trigonometric or by exponential form. The alternative is by algebraic rules but it is really a less effective way in this case. – user Jul 23 '18 at 12:19
  • If you convert the quantity to polar form into a number $r(cos x + i sin x)$ then the real part is the x coordinate and imaginary part is y coordinate. – Shrey Joshi Jul 23 '18 at 12:20

5 Answers5

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For the factor of $(1 - i)^9,$ I think the de Moivre form yields some insight, because $1 - i = \sqrt2 e^{-i\pi/4}.$ Hence $$(1 - i)^9 = 2^{9/2} e^{-i9\pi/4} = 16\sqrt2 e^{-i\pi/4} = 16 - 16i.$$

For the factor of $(3-2i)^3$ I am not convinced by the other answers that computing $\cos(3\arctan(-2/3))$ is simpler than just doing two complex multiplications by the algebraic method.

David K
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If we want proceed in cartesian form we need to expand

$$(3-2i)^3\cdot(1-i)^9=(3-2i)\cdot (3-2i)\cdot(3-2i)\cdot (1-i)\cdot \ldots \cdot (1-i)$$

The best way is by exponential form

$$z=x+iy =re^{i\theta}\implies z^n=r^ne^{in\theta}$$

with

  • $r=\sqrt{x^2+y^2}$
  • $\theta = \arctan (y/x)\, [+\pi]$ (for $x\neq 0$)
user
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  • The exponential form is not the best way. Because the solution is integer, whereas you will end-up with an ugly non-simplifiable trigonometric expression. –  Jul 23 '18 at 12:28
  • @YvesDaoust Excluding the trigonometric form, which way would you suggest? – user Jul 23 '18 at 12:31
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Hint: $(1-i)^2=-2i$ then $(1-i)^8=16$. then $$(3-2i)^3(1-i)^9=16\color{blue}{(3-2i)^2}\color{red}{(3-2i)(1-i)}$$

Nosrati
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2

You can use the binomial formula and exploit simplifications.

$$(3-2i)^3=3^3-3\cdot3^2\cdot2\,i-3\cdot3\cdot2^2+2^3\,i=-9-46i,$$

$$(1-i)^9=(1-i)(1-i)^8=(1-i)(-2i)^4=16(1-i).$$

But in general, there is no shortcut.

Algorithmically speaking, the most efficient way to raise to a high power is by successive squarings, which is better than repeated multiples, but for moderate exponents, you won't see a big difference.

I don't think that the trigonometric form is of much use because

$$(\sqrt{13})^3(\sqrt2)^9e^{-i(3\arctan2/3+9\pi/4)}$$

doesn't tell you much more than $(3-2i)^3(1-i)^9$.

1

If you use DeMoivre's Formula and get $r(\cos x + i\sin x)$ then simply distribute the $r$ to get $$r\cos x + (r\sin x)i$$ Now just evaluate the cosine and sine function and you're left with a complex number in $a+bi$ form.

RayDansh
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