0
$\begingroup$

I've just began studying complex numbers for the first time and there are no solutions to the book I'm following. Are my answers correct for the following introductory questions? Thanks.

Q. Express the following in the form $re^{i\theta}$.

i. $i^3$

Ans: $e^{\frac{\pi i}{2}} $.

ii. $1-i$

Ans: $\sqrt{2} e^{\frac{7\pi i}{4}} $.

iii. $\sqrt{2}(1+i)$

Ans: $2e^{\frac{\pi i}{4}}$.

$\endgroup$
3
  • $\begingroup$ How did you arrive at these answers? Show your work, please. $\endgroup$
    – Yuval Gat
    Jan 26, 2019 at 13:32
  • $\begingroup$ Apologies. I'm using my phone and couldn't get some formatting correct. I took the modulus of the complex number in question to get r, and in order to obtain $\theta$ I used the unit circle . $\endgroup$
    – user606466
    Jan 26, 2019 at 13:36
  • $\begingroup$ While we prefer fractions to be written $\frac nd$ in most places, that format doesn’t work well in an exponent. I would write $e^{\pi i/4}$ rather than $e^{\frac{\pi i}{4}}.$ $\endgroup$
    – David K
    Jan 26, 2019 at 14:42

2 Answers 2

Reset to default
0
$\begingroup$

Yes you are correct on the bottom two.

For the the top one, remember $i^3=-i$

Be careful in questions where they restrict the values of $\theta$. For example if $-\pi <\theta \leq\pi$, your second would be $\sqrt2e^{\frac{-i\pi}{4}}$ instead.

But yeah, just follow the steps I suspect you have been.

If your number is $x+iy$ we have:

$$r=\sqrt{x^2+y^2}$$ $$\theta=\arctan(\frac yx)$$ With the second, double check you got the right angle, and add or subtract $\pi$ to get the correct angle if need be.

$\endgroup$
0
$\begingroup$

For (i), recall that $i = e^{i\pi/2}$. Then $i^3 = (e^{i\pi/2})^3 = e^{3\pi i/2} = e^{-\pi i/2}$.

The other answers are correct.

$\endgroup$

You must log in to answer this question.