Prometheus wrote:

Doing a remedial maths course and i just want to check this question.

I need to state the real part $ \displaystyle x$ and the imaginary part $ \displaystyle y$ of:

$ \displaystyle Z=(-i)^9$

I assume they are expecting me to do it by inspection given it's only worth 1 mark, in which case i make it

$ \displaystyle x=0$ and $ \displaystyle y=-1$ since it's already in Cartesian form.

But i wanted to prove it to myself to make sure i understand the concept not just a mechanical application. So i changed it to polar exponential form in order to expand the brackets and i just want to make sure the workings are correct:

$ \displaystyle r=\sqrt0^2+\sqrt-1^2=1$

$ \displaystyle \theta=tan^{-1} \frac{-1}{0}= -\frac{\pi}{2}$

Is the latter correct, even though it is technically undefined, when thinking of the unit circle it equates to $ \displaystyle \frac{\pi}{2}$?

If so, then $ \displaystyle (e^{-i\frac{\pi}{2}})^9=e^-{i\frac{9\pi}{2}}$

and putting back into Cartesian from:

$ \displaystyle cos(-\frac{9\pi}{2})+isin(-\frac{9\pi}{2})$

Which recovers $ \displaystyle Z=0-i1$

Thanks

Edit: - not +

That is correct, but an awfully hard way to do it.

It is a bit easier to note that $ \displaystyle -i^n = -1^ni^n$ and hence $ \displaystyle -i^9 = -1^9i^9= -1 i^8i = -1(i^2)^4i = -1(-1)^4i = -i$

So the real part is 0 and the imaginary part is -1.