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$Z^2 = \frac{2+5i}{-3+7i}$

I times bottom and top by

${-3-7i}$

coming to

$Z^2=\frac{29-29i}{58} $

ultimately breaking down to

$Z^2= 1/2 +1/2i $

$r^2e^(2i\theta)$ ( i dont know how to put 2 i theta as a power)

for R i got =

$ r^2 = \frac{\sqrt 2}{2}$

$\theta = \frac{\pi}{8}, -\frac{7\pi}{8}$

some help please im sure i did something wrong :(

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1 Answer 1

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There is a minor mistake in your simplification of $\dfrac{2+5i}{-3+7i}$; it is equal to $$\dfrac12-\dfrac i2=\dfrac1{\sqrt2}\left(\cos\left(\frac{7\pi}4\right)+\sin\left(\frac{7\pi}4\right)i\right).$$Therefore,$$Z=\pm\frac1{\sqrt[4]2}\left(\cos\left(\frac{7\pi}8\right)+\sin\left(\frac{7\pi}8\right)i\right).$$

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  • $\begingroup$ for all the values for the principal argument what would be the other answer? $\endgroup$
    – dahyunn
    Commented Nov 21, 2018 at 16:27
  • $\begingroup$ I don't understand. There is no other answer. There cannot be, since every non-zero complex number has exactly two square roots. $\endgroup$ Commented Nov 21, 2018 at 16:40
  • $\begingroup$ the question is asking for 2 values for theta, thanks for helping $\endgroup$
    – dahyunn
    Commented Nov 21, 2018 at 16:50
  • $\begingroup$ @dahyunn Then take$$\frac1{\sqrt[4]2}\left(\cos\left(\frac{7\pi}8\right)+\sin\left(\frac{7\pi}8\right)i\right)\text{ and }\frac1{\sqrt[4]2}\left(\cos\left(\frac{3\pi}8\right)+\sin\left(\frac{3\pi}8\right)i\right).$$ $\endgroup$ Commented Nov 21, 2018 at 16:56

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