Express $-1+i$ in exponential form.
My attempt so far
Let $z=-1+i$
$$r=|z|=\sqrt2$$
$$\theta=\tan^{-1}(-1)=-\frac{\pi}{4}$$
Now, this is where I go wrong (I don't know why it's wrong!):
So in exponential form: $-1+i=\sqrt2 e^{-i\pi/4}$
According to the solutions, $\theta=3\pi/4$.
Thanks in advance
Best Answer
If $\tan \theta = -1$, then
$\cos\theta > 0, \text{ and } \sin\theta <0 \implies \theta = \frac {3\pi}4 $,
or else
$\cos\theta < 0 ,\text{ and } \sin \theta > 1\implies \theta = \frac {7\pi}4 = -\frac{\pi}{4}$.
Since we are working with $-1 + i$, $\cos \theta \lt 0$, $\sin \theta > 0$, and hence, $\theta = \frac{3\pi}{4}$.
Hence, $-1 + i = \sqrt 2e^{3\pi/4}.$