2

On the complex plane , the angle of $i = \pi / 2$ and the angle of $i^2 = \pi$ .

I understand that by definition $i^2 = -1$ but do not understand how to arrive at angle $\pi$ from $\pi / 2$ when we square $i$ on the complex plane. Can anybody help?

Michael Hardy
  • 1
NiallJG
  • 123

7 Answers7

6

You should visualize the complex numbers in the complex plane, that is a number $z = \alpha + \beta i$ has the coordinate $(\alpha,\beta)$ in the complex plane. The angle is then measured from the first axis.

This means in particular that $i = 0+1i$ has the coordinate $(0,1)$, and $i^2 = -1+0i$ has the coordinate $(-1,0)$. These have angles $\frac\pi2$ and $\pi$, respectively.

enter image description here

Eff
  • 12,989
  • 1
    Thanks for the answer, I accept that it is the case and can visualize the answer but I just wasn't able to see how you could get the angle from first principles, everybody's answers were great but KRS got to the crux for me – NiallJG May 22 '15 at 01:14
3

When you multiply complex numbers, their arguments (the angles) add. It is their modules that are multiplied.

Bernard
  • 175,478
2

Multiplying any complex number by $i$ will rotate it 90 degrees counterclockwise. Multiplying any number by $-1$ will rotate it 180 degrees counterclockwise (because this is not different than multiplying by $i$ twice!).

Jonathan Hebert
  • 5,272
2

Any complex number $z$ with magnitude $r$ and angle $\theta$ can be represented in polar form as $z = r(\cos(\theta) + i\sin(\theta))$.

To see what happens when you multiply complex numbers, suppose $z = r(\cos(\theta) + i\sin(\theta))$ and $w = s(\cos(\zeta) + i\sin(\zeta))$.

Then $zw = rs[(\cos(\theta) + i\sin(\theta))(\cos(\zeta) + i\sin(\zeta))] = rs[(\cos(\theta)\cos(\zeta) - \sin(\theta)\sin(\zeta)) + i(\sin(\theta)\cos(\zeta) + \cos(\theta)\sin(\zeta))] = rs[\cos(\theta + \zeta) + i\sin(\theta + \zeta)]$

So the product of two complex numbers $zw$ has an angle $\theta + \zeta$ equal to the sum of the angles of the individual complex numbers $z$ and $w$ that produce it.

In particular, the angle of $i^2$ is $\pi = \frac{\pi}{2} + \frac{\pi}{2}$.

yeons
  • 295
0

Yes the angle of ${i}^{2}$ is $\pi$ bacause -1 is on the left side of the real axis.

Ahmed Abrous
  • 179
0

$z = \alpha + i\beta$

$z(\theta) = x$

$iz(\theta) = x + 90^{\circ}$

$i(\theta) = \dfrac{\pi}{2}$

$i^2(\theta) = \dfrac{\pi}{2} + 90^{\circ} = \dfrac{\pi}{2} + \dfrac{\pi}{2} = \pi$

Jimmy360
  • 649
0

The number of rotations on your complex protractor is written to base $i$ as an exponent at top in terms of $\pi/2 $ radians or 90$^0.$

You have written 2, so it makes $\pi$.

If you had written $ \frac {2}{\pi} $ that would make for one radian as,...

$ i^{2/\pi} = 1. $

Narasimham
  • 40,495