Phasor representation of complex numbers

Instead of typing (3+4i)*(i), how can I multiply two complex numbers in phasor form, in this case

(5∠(arctan(4/3))*(∠90), and convert the result to rectangular form? I have tried representing the complex numbers as [r theta], but typing [5 arctan(4/3)]*[1 90] does not give the correct answer.

Answer 1

[Star Strider]

Converting to phasor representation is definitely taking the long way round.

If you must:

r2p = @(x) [abs(x) rad2deg(angle(x))];                      % Rectangular -> Phasor
p2r = @(x) x(1)*exp(1i*deg2rad(x(2)));                      % Phasor -> Rectangular 
pm = @(x,y) [x(1)*y(1) x(2)+y(2)];                          % Phasor Multiply
pd = @(x,y) [x(1)/y(1) x(2)-y(2)];                          % Phasor Divide

x = 3+4i;
xp = r2p(x);
yp = [1 90];
xptimesyp = pm(xp,yp)
xrtimesyr = p2r(xptimesyp)
Check = x * p2r(yp)

producing:

xptimesyp =
            5       143.13

xrtimesyr =
           -4 +          3i

Check =
           -4 +          3i

.

[Aleem Andrew]

Thank you for your answer

[Star Strider]

As always, my pleasure!

Answer 2

[wahidin syahrir]

thank you very much for your phasor equation sir/miss.

[Star Strider]

My pleasure!