real and imaginary part of complex number

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!

[madhan ravi]

How about:

assume(0<n<1) % after the first line 

[Bruno Luong]

btw

imag (1i^n) 

is

sin(pi*n/2)

[Niclas]

@madhan ravi: thanks for the input , but Matlab still can't solve it this way.

@Bruno Loung: thank you for the solution. But is there no possibility how Matlab can calculate such problems? I'm interested in the way to solve it.

[Torsten]

Write

1i = exp(1i*pi/2) 

and use

exp(1i*pi/2)^n = exp(1i*n*pi/2)
exp(1i*x) = cos(x)+1i*sin(x)

.

[Bruno Luong]

I'm interested in the way to solve it.

Just like Torsen's development. So you'll see that is true for any n that is real, the assumption that (0<n<1) is irrelevant.

[Niclas]

true. Torsten helped me with the understanding, thanks.

Now I get the way how to do it with a pencil. But what if the complex number gets much more complicated?

I still searching for a way to split real and imaginary part with Matlab.

[Torsten]

z^n = (r*exp(i*phi))^n = r^n * (cos(n*phi) + i*sin(n*phi))

Re(z^n) = r^n*cos(phi*n)
Im(z^n) = r^n*sin(phi*n)

[Niclas]

z = r^n*(cos(n*phi)+1i*sin(n*phi))
Re = real(z)
Im = imag(z)

gives in my Matlab

Re =
 
real(r^n)*cos(n*phi) - imag(r^n)*sin(n*phi)
 
 
Im =
 
imag(r^n)*cos(n*phi) + real(r^n)*sin(n*phi)

these are obviously not the solutions you wrote and I'm looking for.

[Torsten]

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]

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?

Answer 1

[Walter Roberson]

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');

[Niclas]

works perfect for this function, thanks!

I used it without the data to work on.

can I ask you a subsequent question? Why does this not work for (i*x)^n

syms x real
syms n real
Vn = ((1i*x)^n);
RV = rewrite( real(Vn), 'exp')
IV = rewrite( imag(Vn), 'exp')

I expected something like

RV=x^n*sin(pi/2*n)
IV=x^n*cos(pi/2*n)

but Matlab gave:

RV =real(exp(n*log(a))*exp((pi*n*1i)/2))
 
 
IV =imag(exp(n*log(a))*exp((pi*n*1i)/2))

[Walter Roberson]

Current versions of MATLAB give

RV =

exp(n*log(abs(x)))*cos(n*angle(x*1i))

IV =

exp(n*log(abs(x)))*sin(n*angle(x*1i))

If you can constrain x further then it can produce better results:

syms x positive
syms n real
Vn = ((1i*x)^n);
RV = rewrite( real(Vn), 'exp')
IV = rewrite( imag(Vn), 'exp')

Either way you might want to simplify(RV) and simplify(IV)

[Niclas]

really nice Walter, this works perfectly!

How about going one step further: it does not work with (1+(1i*x)^n)^y

syms x positive
syms y positive
syms n real
Vn = (1+(1i*x)^n)^y;
RV = simplify(rewrite( real(Vn), 'exp'))
IV = simplify(rewrite( imag(Vn), 'exp'))

I expected something like (take care, this is wrong!)

RV=1+x^n*y*cos(pi*n*y/2)
IV=x^n*sin(pi*n*y/2)

but only got

RV =real(((-1)^(n/2)*x^n + 1)^y)
IV =imag(((-1)^(n/2)*x^n + 1)^y)

. Is there an other possibility to expand this to get a solution like before?

[Torsten]

Do you know of an expansion for (1+x)^y if x and y are real positive numbers except an infinite Taylor series ? I don't.

[Niclas]

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.

[Torsten]

y positive integer

might work.