real and imaginary part of complex number

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Niclas
Niclas 2019-7-15
评论： Torsten 2019-7-16
Hi,
I'm trying to get the real and imaginary part of a formular in a limited range of a parameter n.
how do I add in my formular
syms n real
imag (1i^n)
that 0 < n < 1 ?
Thanks a lot!
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Torsten
Torsten 2019-7-15
r is real - it is the length of the vector from the origin to z.
Thus
real(r^n) = r^n, imag(r^n) = 0.
Niclas
Niclas 2019-7-16
yes, the two solutions are equal.
But how can I use Matlab to transform the equation?
If I give Matlab
imag(1I^n)
as complex number, why the program don't know how to transform it to polar coordinates? And why Matlab don't get from that
imag(r^n) = 0
?
If there any possibility to not do this transformation manually?

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Walter Roberson
Walter Roberson 2019-7-16
V = complex(randi([-10 10],1), randi([-10 10])); %some data to work on
syms n real
Vn = V^(1/n);
RV = rewrite( real(Vn), 'exp');
IV = rewrite( imag(Vn), 'exp');
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Niclas
Niclas 2019-7-16
Good point Torsten.
I guess we need constrain y further. So why dont take the assumption from begin:
assume(0<y<1)
The Matlab solution stays the same.

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