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I'm trying to get the exact answer to the following complex expression, with the approximate value being $-.12256 + .74486i$ (according to WolframAlpha). It looks deceptively simple, but I don't think my calculus-level education can get the job done:

$0\:=z+1-iz^{-\frac{1}{2}}$

I tried changing it to a different form (this is the form I want to get the answer in), based on Euler's identity:

$0\:=r\left(-1\right)^{\theta }+1-ir\left(-1\right)^{-\frac{\theta }{2}}$

or even:

$0\:=r\left(-1\right)^{\frac{\theta }{\pi }}+1-ir\left(-1\right)^{-\frac{\theta }{2\pi }}$

Can this be done?

Ibrahim Mahmoud
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1 Answers1

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You can't change it to a different form based on Euler's Identity, therefore the form you're using is incorrect. In fact, there is an easier way to solve this:

$$ z + 1 -iz^{-1/2} = 0$$ $$ <=> z + 1 = iz^{-1/2} $$ $$ <=> (z + 1)^{2} = (iz^{-1/2})^{2} $$ $$ <=> z^{2} + 2z + 1 = -z^{-1} $$ $$ <=> z^{2} + 2z + 1 + z^{-1} = 0 $$ $$ <=> z^{3} + 2z^{2} + z + 1 = 0 $$

Solving this equation, you will get three different approximate roots: $-1.754877...$, $−.12256+.74486i$ and $−.12256-.74486i$

About why your form is wrong, let's consider an example with $z = \frac{5}{2} + \frac{5 \sqrt{3}}{2}i$.In this example, z has $r = 5$ and $\theta = \frac{\pi}{3}$. Writing this with the form $ z = re^{i \theta} $ and using Euler's Identity we get:

$$ z = 5e^{i \frac{\pi}{3}} $$ $$ = 5(e^{i \pi})^{\frac{1}{3}} $$ $$ = 5(-1)^{\frac{1}{3}} $$ $$ = -5 $$ which contradict to the real value of z above.

The 2nd
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  • I've been thinking about this for a while, because it makes sense yet something seemed off from my understanding. I think the difference relies with how exponentiation is done, I was expecting complex exponentiation rather than discrete real exponentiation (https://math.stackexchange.com/questions/317528/how-do-you-compute-negative-numbers-to-fractional-powers). Plugging your example into Wolfam Alpha, I get the desired result: "https://www.wolframalpha.com/input/?i=5*%28-1%29%5E%281%2F3%29". – Ibrahim Mahmoud Mar 12 '20 at 23:15