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I'm having trouble with the following:

$(-16i)^{5/4}$

My calculations for the Principal root is:

$32(\cos (3\pi/2) * 5/4) + i \sin (3\pi/2)* 5/4))$

$=32(Cis (15\pi/8))$

This answer does not agree with the online calculators. It gives a positive real value and the online calculators show a principal angle in Quadrant 3.

Confused on what happened here.

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3 Answers 3

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$$z=(-16i)^{5/4}=32(-i)^{5/4}$$

Use

$$-i=cis\left(\frac{3\pi}{2}+2k\pi\right)$$

Then

$$(-i)^{5/4}=(-i)^{1/4}=cis\left(\frac{3\pi}{8}+\frac{1}{2}k\pi\right)$$

choose $k \in \{0,1,2,3\}$. So the solutions will be:

$$k=0 \rightarrow z_0=32\cdot cis\left(\frac{3\pi}{8}\right)\\ k=1 \rightarrow z_1=32\cdot cis\left(\frac{7\pi}{8}\right)\\ k=2 \rightarrow z_2=32\cdot cis\left(\frac{11\pi}{8}\right)\\ k=3 \rightarrow z_3=32\cdot cis\left(\frac{15\pi}{8}\right) $$

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  • $\begingroup$ What happens to the -16? $\endgroup$
    – user163862
    Commented Jan 11, 2017 at 19:42
  • $\begingroup$ the result will be multplied by $32$ like you already did. $\endgroup$
    – Arnaldo
    Commented Jan 11, 2017 at 19:43
  • $\begingroup$ $k$ is 0 , 1 , 2 , 3 $\endgroup$
    – Nosrati
    Commented Jan 11, 2017 at 19:45
  • $\begingroup$ I completely don't understand this. So when k=0 my answer would be $32Cis(3\pi/8)$? $\endgroup$
    – user163862
    Commented Jan 11, 2017 at 19:46
  • $\begingroup$ @user163862: yes it is! $\endgroup$
    – Arnaldo
    Commented Jan 11, 2017 at 19:48
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Let $z=-16i$ and $n=\dfrac54$. For solving you have to compute $r=|z|$ and argument $\theta$ where $\tan\theta=\dfrac{y}{x}$.

then $r=|z|=|-16i|=16$ and argument $\theta=\dfrac{3\pi}{2}$. Then write $$z_k=r^n(\cos n\theta+i\sin n\theta)$$

But argument adds with $2k\pi$ so we have $$z_k=r^n\Big(\cos n(\theta+2k\pi)+i\sin n(\theta+2k\pi)\Big)$$ so $$z_k=16^\dfrac54\Big(\cos\dfrac54(\dfrac{3\pi}{2}+2k\pi)+i\sin\dfrac54(\dfrac{3\pi}{2}+2k\pi)\Big)$$ and for $k=0,1,2,3$ write $$z_k=32\Big(\cos(\dfrac{15\pi}{8}+\frac{5k\pi}{2})+i\sin(\dfrac{15\pi}{8}+\frac{5k\pi}{2})\Big)$$ finally for $k=0,1,2,3$ we conclude that \begin{eqnarray} k=0 &\Rightarrow& z_0=32\Big(\cos\dfrac{15\pi}{8}+i\sin\dfrac{15\pi}{8}\Big)=19.28+25.5i\\ k=1 &\Rightarrow& z_1=32\Big(\cos\dfrac{35\pi}{8}+i\sin\dfrac{35\pi}{8}\Big)=12.24+29.5i\\ k=2 &\Rightarrow& z_2=32\Big(\cos\dfrac{55\pi}{8}+i\sin\dfrac{55\pi}{8}\Big)=-29.5+12.24i\\ k=3 &\Rightarrow& z_3=32\Big(\cos\dfrac{75\pi}{8}+i\sin\dfrac{75\pi}{8}\Big)=-12.24-29.5i \end{eqnarray} This was step by step solving.

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  • $\begingroup$ The exact values of trigonometric functions of multiples of $\frac \pi 8$ are known. Have a look at mathworld.wolfram.com/TrigonometryAnglesPi8.html $\endgroup$ Commented Jan 12, 2017 at 6:12
  • $\begingroup$ @ClaudeLeibovici Thanks. Numerical values are about user and I'm not interested to numerical solutions. $\endgroup$
    – Nosrati
    Commented Jan 12, 2017 at 6:20
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First write $-16i$ in general polar form: $16e^{3\pi i/2+2\pi ik}$. Then $$(16e^{3\pi i/2+2\pi ik})^{5/4}=32e^{15\pi i/8+5\pi ik/2}=32e^{15\pi i/8+20\pi ik/8}.$$

By playing with values of $k$, we obtain the separate roots $32e^{\frac{3+4k}{8} \pi i}$.

Now, the discrepancy here is which of these is actually the principal root. I believe this is a matter of definitions. According to Wikipedia, the principal root is the one with the smallest positive argument. By that definition the answer has argument $3\pi/8$. However, I believe that Wolfram (and your calculator) has calculated the principal root by beginning with the smallest absolute value argument, which gives $16e^{-\pi i/2}$, and raising to the $5/4$, giving $32e^{-5\pi i /8}=32e^{11\pi i/8}$, which I believe is the answer you are looking for.

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  • $\begingroup$ Thank you. It's easy to see now that the principal root is not set in stone. $\endgroup$
    – user163862
    Commented Jan 23, 2017 at 19:26

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