# Express $-1+i$ in exponential form.

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, $\theta=3\pi/4$.

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@MrCroutini If a picture is good enough for you, then draw one,locate the point $(-1,1)\sim -1+i$ and find the angle. – Git Gud Jan 5 '14 at 13:00
@MrCroutini $(-1,1)$ isn't in the bottom left quadrant. – Git Gud Jan 5 '14 at 13:04
Where does $\pi/3$ come from? You had $\sqrt{2}e^{-i\pi/4}$, which in fact gives you $1-i$. The angle $-\pi/4$ points in the lower half of the right half plane. The point you're looking for is in the upper half of the left half plane, it is the negative of what you had, so you multiply with $-1$, which is adding $\pi$ to the angle. – Daniel Fischer Jan 5 '14 at 13:06
Well, the formulation isn't right, the angle "is" not in any quadrant, it's the ray determined by the angle that is in the quadrant, or we can say the angle "points into" the quadrant. But the idea is right. – Daniel Fischer Jan 5 '14 at 13:11
@MrCroutini Indeed, "right" in that comment was used as a synonym of "correct". Unfortunate choice of word, sorry. – Daniel Fischer Jan 5 '14 at 13:13

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, $-1 + i = \sqrt 2e^{3\pi/4}.$

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Needs a TU! ... +1 – Amzoti Jan 5 '14 at 14:37
@amWhy: Nope! :-) But I feel there is a match or something like a playoff game on the way. Isn't it? – Babak S. Jan 5 '14 at 15:54
@amWhy: It seems to me that you are stick to one side and are eager to see they win the title. I am watching the countdown on its page right now. I vote for Packers not 49ers. – Babak S. Jan 5 '14 at 16:04

You almost got it! $$-1+i=\sqrt{2}\,\mathrm{e}^{3\pi i/4}=\mathrm{e}^{\ln 2/2}\,\mathrm{e}^{3\pi i/4}=\ldots$$

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I think it's the angle that's the issue. – Mr Croutini Jan 5 '14 at 12:55
@GitGud: Thnx - Corrected it! – Yiorgos S. Smyrlis Jan 5 '14 at 13:00
@YiorgosS.Smyrlis This doesn't help the OP, though. He doesn't understand why it should be $3\pi /4$. – Git Gud Jan 5 '14 at 13:01