The following diagram appears in my textbook: enter image description here

They follow with enter image description here

I am unsure of this visualization as I've learnt that the conjugate of a complex number in the form $$z=x+yi,x,y,\in\mathbb{R}$$ is simply $$z^*=x-yi.$$ This would mean, in polar form,

$$\DeclareMathOperator\cis{cis} z^*=r\cis(-\theta).$$

Is there something I am missing here? I'm genuinely confused as it seems that sources online disprove what is claimed in my textbook, yet my textbook continues to use this definition for later questions.

  • $\begingroup$ This textbook is not using standard conventions: $z^*$ and $-z^*$ ought to be switched, assuming that you measure angles counterclockwise: from the positive real (horizontal) axis towards the positive imaginary (vertical) axis. $\endgroup$ Mar 10, 2021 at 3:55
  • $\begingroup$ But also note that $-\theta$ and $2\pi - \theta$ differ by a complete rotation $(2\pi)$ so they're interchangeable as angles: $\operatorname{cis}(-\theta) = \operatorname{cis}(2\pi-\theta)$. $\endgroup$ Mar 10, 2021 at 3:58
  • $\begingroup$ Thanks for the clarification! I take it that it is safe to say that $-z^*$ is not $r cis(-\theta)$? $\endgroup$
    – Sean Xie
    Mar 10, 2021 at 4:02

2 Answers 2



  • $r\,\mathrm{cis}(\pi+\theta),$
  • $r\,\mathrm{cis}(\pi-\theta),$ and
  • $r\,\mathrm{cis}(2\pi-\theta),$

only the third one refers to the complex conjugate of $r\,\mathrm{cis}(\theta),$ according to the standard definition.

Assuming that in this textbook $\mathrm{Re}(z)$ is indeed along the $x$-axis (i.e., $z=x+iy$), we can infer that the author

  1. uses $(0,2\pi]$ as the principal argument, and
    • either writes $z$'s conjugate as $(-z^*),$

    • or writes $z$'s conjugate as $(z^*)$ but defines it as $(-x+iy).$

(1.) is fine (both $(-\pi,\pi]$ and $(0,2\pi]$ are conventional), but (2.) is strange and highly nonstandard.


That's taken from the IB Oxford AA HL textbook. In IB, the real scale is the x-axis and the imaginary scale is the y-axis. According to that, the book is wrong. It has gotten the conjugate and the conjugate of the opposite the wrong way around.

  • $\begingroup$ That's correct, I was an IB student at the time and I was quite confused by the error. Hopefully, they've revised it in the next editions :) $\endgroup$
    – Sean Xie
    Jan 29 at 3:50

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