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Question: what is recommended way to find magnitude of complex number?

Question:what is recommended way to find magnitude of complex number?

nm 8552 Maple 2021
radical simplification trig abs
January 16 2022
1

I read that abs(v) should find the magnitude of complex number. But I find cases where it does not.

Should one then use sqrt( Re(v)^2 + Im(v)^2 ) instead or is there different command to use?  Here is an example

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interface(version)

`Standard Worksheet Interface, Maple 2021.2, Windows 10, November 23 2021 Build ID 1576349`

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restart;
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v:=1/16*(I*5^(1/2)+I-4*sin(1/5*Pi))*2^(3/10)*(5-5^(1/2))^(1/2)-1/8*2^(4/5)*((I*5^(1/2)+I)*sin(1/5*Pi)+1/2*5^(1/2)+3/2)

(1/16)*(I*5^(1/2)+I-4*sin((1/5)*Pi))*2^(3/10)*(5-5^(1/2))^(1/2)-(1/8)*2^(4/5)*((I*5^(1/2)+I)*sin((1/5)*Pi)+(1/2)*5^(1/2)+3/2)

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abs(v)

-(1/16)*(I*5^(1/2)+I-4*sin((1/5)*Pi))*2^(3/10)*(5-5^(1/2))^(1/2)+(1/8)*2^(4/5)*((I*5^(1/2)+I)*sin((1/5)*Pi)+(1/2)*5^(1/2)+3/2)

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sqrt( Re(v)^2 + Im(v)^2 );

((-(1/4)*2^(3/10)*(5-5^(1/2))^(1/2)*sin((1/5)*Pi)-(1/8)*2^(4/5)*((1/2)*5^(1/2)+3/2))^2+((1/16)*2^(3/10)*(5-5^(1/2))^(1/2)*(5^(1/2)+1)-(1/8)*2^(4/5)*(5^(1/2)+1)*sin((1/5)*Pi))^2)^(1/2)

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Download complex.mw

Here is the same thing in Mathematica:

I searched help for modulus but could not find one.

Maple 2021.2
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