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Multiplication of non-zero complex number with $~-i~$ rotates the point about origin through a right angle in anticlockwise direction. True or false?

My attempt :- actually I don't know how to proceed in general but I did it with an example of $~1+i~$ and by multiplying it with $~-i~$ the point turns $~ 90~$ in clockwise direction. Also if we multiple the above complex number by $~i~$ it rotates by $~90~$ in anti-clockwise direction. Is the above result is true in general of just for particular example.

Also general proof will be appreciated. Thanks

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    $\begingroup$ OK, not everybody can be bothered to switch to upper case when typing the word I. But surely when you are talking about multiplication by the square root of $-1$, you can seen the sense in making the effort? $\endgroup$
    – TonyK
    Commented Jul 27, 2019 at 13:22
  • $\begingroup$ Yes, sorry i forget about it $\endgroup$
    – Adesh Sharma
    Commented Jul 28, 2019 at 11:16
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    $\begingroup$ It seems you are incorrigible :-) $\endgroup$
    – TonyK
    Commented Jul 28, 2019 at 11:18

3 Answers 3

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I assume that by “anticlockwise” you mean “with the same verse of rotation by which the positive $x$-semiaxis reaches the positive $y$-axis by the least angle”; “clockwise” is the opposite verse of rotation.

Then multiplication by $-i$ induces a clockwise rotation by a right angle, because $1\cdot(-i)=-i$, which in the standard representation of complex numbers has a negative $y$-coordinate.

As you see, there are many implicit assumptions in the statement of the problem.

If you decide to represent the $x$-axis in the usual fashion (horizontal, positive verse from left to right) and the $y$-axis upside down with respect to the usual way, then, with the intuitive understanding of clockwise and anticlockwise, multiplication by $-i$ induces an anticlockwise rotation by a right angle. That's why I initially gave a definition that's independent on the graphical representation.

Another case: suppose your teacher is drawing on a side of a transparent board and you're on the opposite side; you and the teacher will see different verses of rotation, under the common understanding of clockwise ad anticlockwise. Not if you stick to the definition I gave.

By the way, the “anticlockwise” verse I defined at the top is usually and better called positive.

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It's easy to see in general, by drawing in the coordinate system. The complex number $a+bi$ is represented as (the vector from the origin to) the point with coordinates $(a, b)$.

We have $$(a+bi)\cdot i=-b+ai\\ (a+bi)\cdot(-i)=b-ai$$

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  • $\begingroup$ So then what can we conclude now? $\endgroup$
    – Adesh Sharma
    Commented Jul 28, 2019 at 11:22
  • $\begingroup$ That multiplying by $\pm i$ rotates the vector corresponding to the complex number by $\pm 90^\circ$. $\endgroup$
    – Berci
    Commented Jul 28, 2019 at 13:56
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This is wrong. Usually one puts the unit $i$ on the vertical coordinate axis in upright position, so that to go from $1$ to $i$ is an anti-clockwise rotation by $90°$. Then $-i$ is in the downward position and multiplication by $-i$ rotates by $90°$ in clockwise direction. Remember that $1=1+0i$ is at the 3:00 position on the clock, $i$ at the 12:00 position and $-i$ at the 6:00 position.

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  • $\begingroup$ According to you also multiplying by '-i' rotate the complex number in clockwise direction, then why so said this is wrong! $\endgroup$
    – Adesh Sharma
    Commented Jul 28, 2019 at 11:44
  • $\begingroup$ Because in your question, the first claim is that the rotation by $-i$ is anti-clockwise. $\endgroup$
    – Lutz Lehmann
    Commented Jul 28, 2019 at 13:09

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