I am just starting with complex numbers and vectors. The question is:

Convert the following to Cartesian form.

a) $8 \,\text{cis} \frac \pi4$

The formula given is:

$$z = x +yi = r\space(\cos\theta + i\sin\theta)$$

With $r=8$ and $\theta = \frac\pi4$, I did:

$$z=8\left(\cos\frac\pi4 + i \sin \frac\pi4\right)$$ $$z = 8(0.71 + i0.71)$& $$z = 5.66 + i5.66$$

The answer they give in the answers section is:


I know this is the same answer just written approximately. Is someone able to take me through how you get to $z=4\sqrt2(1+i)$ instead of $z = 5.66 + i5.66$?

  • $\begingroup$ $\cos\dfrac\pi4=\dfrac1{\sqrt2}$ $\endgroup$ – lab bhattacharjee Feb 27 '18 at 4:05
  • $\begingroup$ I know this is the same answer just written differently No, because $\,4 \sqrt{2} \ne 5.66\,$. $\endgroup$ – dxiv Feb 27 '18 at 4:13
  • $\begingroup$ @dxiv Thank you. Should have written approximately. $\endgroup$ – Juan Pablo Feb 27 '18 at 4:15
  • $\begingroup$ Also, while everyone is here, does anyone know why when I write fractions in Mathsjax they come out so small? And everyone elses are normal size. I am new to stack exchange and mathsjax $\endgroup$ – Juan Pablo Feb 27 '18 at 4:16
  • $\begingroup$ @JuanPablo That was the point. When an exact result is readily available, just an approximation won't cut it. P.S. About fractions, try $$\frac{a}{b}$$ or $\dfrac{a}{b}$ or $\displaystyle\frac{a}{b}$. See here for more MathJax tips. $\endgroup$ – dxiv Feb 27 '18 at 4:17

Knowing the $\pi\over4$ family:

$$\cos \dfrac{\pi}{4} = \sin \dfrac{\pi}{4} = \dfrac{1}{\sqrt 2}$$

Then applying this result to $8\,\text{cis} \frac{\pi}{4}$:

$$z=8\left(\dfrac{1}{\sqrt 2} + i\dfrac{1}{\sqrt 2}\right) = 8\cdot \dfrac{1}{\sqrt 2}(1+i)$$

And then multiply:

$$8\cdot \dfrac{1}{\sqrt 2} = \dfrac{8\sqrt 2}{2}=4\sqrt2$$

$$8\cdot \dfrac{1}{\sqrt 2}(1+i) = 4\sqrt 2(1+i)$$

  • $\begingroup$ Ok, that makes sense, I was stuck at the very first bit. Much appreciated. $\endgroup$ – Juan Pablo Feb 27 '18 at 4:12
  • $\begingroup$ @JuanPablo You've got to remember your unit circle :) $\endgroup$ – Andrew Li Feb 27 '18 at 4:13

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