Compute Powers and Put Them in the Form a+bi

Question

$$(-\sqrt3-i)^7$$

I'm not sure how to get the answer? The answer is:

$$(2^6)\sqrt3+(2^6)i$$

sorry about the formatting I'm still new

Answer 1

Follow the steps below,

$$(-\sqrt3-i)^7=\left[2\left(-\frac{\sqrt3}{2}-\frac12i\right)\right]^7$$ $$=2^7\left(\cos\frac{7\pi}{6}+\sin\frac{7\pi}{6}i\right)^7 =2^7(e^{i\frac{7\pi}{6}})^7=2^7e^{i\frac{49\pi}{6}}$$ $$=2^7\left(\cos\frac{49\pi}{6}+\sin\frac{49\pi}{6}i\right)=2^7\left(\cos\frac{\pi}{6}+\sin\frac{\pi}{6}i\right)$$ $$=2^7\left(\frac{\sqrt3}{2}+\frac{1}{2}i\right)=2^6(\sqrt3+i)=2^6\sqrt3+2^6 i$$

Answer 2

As Ninad Munshi wrote, to compute $(a+bi)^n$, first comvert $a+bi$ to polar form.

To do this, want $a+bi =re^{it} $ where $r \ge 0$ is real and $0 \le t \lt 2\pi$.

Then $a+bi =re^{it} =r(\cos(t)+i\sin(t)) =r\cos(t)+ir\sin(t) $ so $a = r\cos(t), b = r\sin(t) $.

Then $a^2+b^2 =r^2(\cos^2(t)+\sin^2(t)) =r^2 $ so $r = \sqrt{a^2+b^2} $.

Then $t$ is gotten by $\tan(t) =\dfrac{r\sin(t)}{r\cos(t)} =\dfrac{b}{a} $ and adjusting the quadrant that $t$ is in depending on the signs of $a$ and $b$.

Finally, if $a+bi = re^{it}$, $(a+bi)^n = (re^{it})^n =r^ne^{int} =r(\cos(nt)+i\sin(nt) $.

Answer 3

Another way is to use the binomial theorem. This is usually harder than the polar form version.

$\begin{array}\\ (a+ib)^n &=\sum_{k=0}^n \binom{n}{k}a^k (ib)^{n-k}\\ &=\sum_{k=0}^n \binom{n}{k}a^{n-k} (ib)^{k}\\ &=\sum_{k=0}^n \binom{n}{k}a^{n-k} i^kb^{k}\\ &=\sum_{k=0}^{\lfloor n/2\rfloor} \binom{n}{2k}a^{n-2k} i^{2k}b^{2k}+\sum_{k=1}^{\lfloor (n+1)/2\rfloor } \binom{n}{2k-1}a^{n-2k+1} i^{2k-1}b^{2k-1}\\ &=\sum_{k=0}^{\lfloor n/2\rfloor} \binom{n}{2k}a^{n-2k} (-1)^{k}b^{2k}+\sum_{k=1}^{\lfloor (n+1)/2\rfloor } \binom{n}{2k-1}a^{n-2k+1} (-i)(-1)^{k}b^{2k-1}\\ &=\sum_{k=0}^{\lfloor n/2\rfloor} \binom{n}{2k}a^{n-2k} (-1)^{k}b^{2k}-i\sum_{k=1}^{\lfloor (n+1)/2\rfloor } \binom{n}{2k-1}a^{n-2k+1} (-1)^{k}b^{2k-1}\\ \end{array} $

Answer 4

As others have pointed out, this problem can be solved using polar coordinates and DeMoivre's formula, mainly because the angle turns out to be a nice fraction of $\pi$. But if you know how to multiply complex numbers, then at the very least you can simply compute a seventh power step by step:

$$\begin{align} x=-\sqrt3-i &\implies x^2=(-\sqrt3-i)(-\sqrt3-i)=2+2\sqrt3 i\\ &\implies x^3=(-\sqrt3-i)(2+2\sqrt3i)=-8i\\ &\implies x^4=(-\sqrt3-i)(-8i)=-8+8\sqrt3i\\ &\implies\,\,etc. \end{align}$$

But of course the nice result $x^3=-8i$ facilitates a short circuit:

$$x^7=x^3x^3x=(-8i)(-8i)(-\sqrt3-i)=-64(-\sqrt3-i)=64\sqrt3+64i$$

(Note, $64=2^6$) It's the nice result for $x^3$ that makes the DeMoivre approach work, and it can be worth spending some time thinking about why that is so. The take-home message is that, if you're asked to do a computation and you can't immediately think of an easy way to do it, just go ahead and start computing, and see what you learn along the way. Even if you wind up not learning anything, six multiplications aren't all that many to do, and at least you've got the answer -- and some useful practice in arithmetic.