Writing $(\sqrt{3}+3i)^{18}$ and $(i-1)^{-11}$ in the form $a+bi$

Question

I have two current math problems I just can't solve. I'm to express the following in the form $a+bi$:

$(\sqrt{3}+3i)^{18} \qquad\text{and}\qquad (i-1)^{-11}$

The answer for the first one is $12^9$; and for the second, it's $\frac{1}{64}(1-i)$.

The results I get keep ending up into something that isn't even worth mentioning.

Answer 1

Hint:

Use the exponential form of the complex numbers:

• The module of $\sqrt 3+3i$ is $\sqrt{12}=2\sqrt3$, hence $$\sqrt 3+3i=2\sqrt3\biggl(\frac12+\frac{\sqrt3}2i\Biggr)=2\sqrt3\,\mathrm e^{\tfrac{i\pi}3}$$ and $\;(\sqrt 3+3i)^{18}=(2\sqrt3)^{18}\,\Bigl(\mathrm e^{\tfrac{6i\pi}3}\Bigr)^3=12^9\,1^3=12^9$.
• Proceed in the same way for $(i-1)^{11}$.

Answer 2

This is a solution to the problem with as few prerequisites as possible. We have $$(\sqrt{3}+3\cdot i)^{18}= \big[(\sqrt{3}+3\cdot i)^{3}\big]^{6}. $$ Note that \begin{align} (\sqrt{3}+3\cdot i)^{2}=&(\sqrt{3}+3\cdot i)\cdot (\sqrt{3}+3\cdot i)\\ =& 3 +6\sqrt{3}\cdot i-9\\ =& -6+6\sqrt{3}\cdot i \end{align} and \begin{align} (\sqrt{3}+3\cdot i)^{3}=&(\sqrt{3}+3\cdot i)\cdot (-6+6\sqrt{3}\cdot i)\\ =& \sqrt{3}\cdot (-6+6\sqrt{3}\cdot i)+3\cdot i\cdot(-6+6\sqrt{3}\cdot i)\\ =& -6\sqrt{3}+18\cdot i-18\cdot i-18\sqrt{3}\\ =& -24\cdot\sqrt{3}\\ \end{align} Now it's up to you. In the same way, do the other exercise.