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Algebra Level pending

Define a complex number z z which satisfied below expression:

z = 1 2 + 1 2 ( 1 2 + 1 2 ( 1 2 + 1 2 ( . . . ) 2 ) 2 ) 2 z=\frac{1}{\sqrt2} +\frac{1}{\sqrt2}(\frac{1}{\sqrt2} +\frac{1}{\sqrt2}(\frac{1}{\sqrt2} +\frac{1}{\sqrt2}(...)^2)^2)^2

and the coefficient of the imagine part of z z is positive.

Find argument of complex number z z in degrees.

Clarification: You may assume that the argument is between 0 ∘ 0^\circ to 36 0 ∘ 360^\circ .

45 60 90 180

Kelvin Hong by Kelvin Hong , 21, Malaysia

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1 solution

Kelvin Hong
Jul 8, 2017

I have thought this problem by some inspiration:

i = c o s ( π / 2 ) + i s i n ( π / 2 ) = c o s ( π / 4 ) + i s i n ( π / 4 ) = 1 2 + 1 2 i \sqrt{i}=\sqrt{cos(\pi/2)+isin(\pi/2)}=cos(\pi/4)+isin(\pi/4)=\frac{1}{\sqrt2}+\frac{1}{\sqrt2}i

Then

i = ( 1 2 + 1 2 i ) 2 i=(\frac{1}{\sqrt2}+\frac{1}{\sqrt2}i)^2

Compute this to first expression, we get

i = 1 2 + 1 2 ( 1 2 + 1 2 i ) 2 = 1 2 + 1 2 ( 1 2 + 1 2 ( 1 2 + 1 2 i ) 2 ) 2 = 1 2 + 1 2 ( 1 2 + 1 2 ( 1 2 + 1 2 ( . . . ) 2 ) 2 ) 2 \sqrt{i}=\frac{1}{\sqrt2}+\frac{1}{\sqrt2}(\frac{1}{\sqrt2}+\frac{1}{\sqrt2}i)^2=\frac{1}{\sqrt2}+\frac{1}{\sqrt2}(\frac{1}{\sqrt2}+\frac{1}{\sqrt2}(\frac{1}{\sqrt2}+\frac{1}{\sqrt2}i)^2)^2=\frac{1}{\sqrt2} +\frac{1}{\sqrt2}(\frac{1}{\sqrt2} +\frac{1}{\sqrt2}(\frac{1}{\sqrt2} +\frac{1}{\sqrt2}(...)^2)^2)^2

So I have proved that z = i z=\sqrt{i}

then a r g ( z ) = 4 5 ∘ arg(z)=45^\circ

Edited at 9/7/2017:

This is proof of question:

We see that

z = 1 2 + 1 2 z 2 z=\frac{1}{\sqrt2}+\frac{1}{\sqrt2}z^2

So

z 2 − 2 z + 1 = 0 z^2-\sqrt2 z+1=0

Solve for this, we get

z = 1 2 + 1 2 i z=\frac{1}{\sqrt2}+\frac{1}{\sqrt2}i or z = 1 2 − 1 2 i z=\frac{1}{\sqrt2}-\frac{1}{\sqrt2}i

By problem statement, choose the first answer,so argument of z z is 4 5 ∘ \boxed{45^\circ}

1 Helpful 0 Interesting 0 Brilliant 0 Confused

Let x n + 1 = 1 2 + 1 2 x n 2 x_{n+1}=\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}{x_n}^2 .

What about the possibility when lim ⁡ n → ∞ x n \lim_{n\rightarrow\infty} x_n diverges?

Boi (보이) - 3 years, 11 months ago

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How can you says a term diverges? Is that complex number also counted in? Actually I found out z z can be expressed in two form: e i π 4 e^{\frac{i\pi}{4}} and e 7 i π 4 e^{\frac{7i\pi}{4}} So, do you find another solution of z z ?

Kelvin Hong - 3 years, 11 months ago

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I meant your solution does not justify the answer.

You know that π 2 > 1 \dfrac{\pi}{2}>1 , and thus e π 2 > 2 e^{\frac{\pi}{2}}>2 .

Consider ( e π 2 ) ( e π 2 ) ( e π 2 ) ( e π 2 ) ⋯ {(e^{\frac{\pi}{2}})}^{{(e^{\frac{\pi}{2}})}^{{(e^{\frac{\pi}{2}})}^{{(e^{\frac{\pi}{2}})}^{\cdots}}}} .

It is obvious that it diverges to infinity.

However, consider e π 2 i = i e^{\frac{\pi}{2}i}=i .

Then, if your method is correct, i = e π 2 i = ( e π 2 ) ( e π 2 i ) = ( e π 2 ) ( e π 2 ) ( e π 2 i ) = ( e π 2 ) ( e π 2 ) ( e π 2 ) ( e π 2 ) ⋯ i=e^{\frac{\pi}{2}i}={(e^{\frac{\pi}{2}})}^{(e^{\frac{\pi}{2}i})}={(e^{\frac{\pi}{2}})}^{{(e^{\frac{\pi}{2}})}^{(e^{\frac{\pi}{2}i})}}={(e^{\frac{\pi}{2}})}^{{(e^{\frac{\pi}{2}})}^{{(e^{\frac{\pi}{2}})}^{{(e^{\frac{\pi}{2}})}^{\cdots}}}} .

This doesn't make sense. So, the answer is not proven right.

Boi (보이) - 3 years, 11 months ago

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@Boi (보이) – Oh, ok, maybe imagine number cannot compare by this way I think. But I have thought a prove to this. I will posted it soon and please check my proof! Thanks for your suggestion!

Kelvin Hong - 3 years, 11 months ago

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@Kelvin Hong – No problem, and ok!

Boi (보이) - 3 years, 11 months ago

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@Boi (보이) – I appreciate your help!

Kelvin Hong - 3 years, 11 months ago

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