Is it possible to perform basic operations on complex numbers in polar form?

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Is it possible to perform basic operations on complex numbers in polar form?

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Answer 1 · 2015-01-06T14:42:33

Yes, of course.

Polar form is very convenient to multiply complex numbers.
Assume we have two complex numbers in polar form:
$z_{1} = r_{1} [cos (ϕ_{1}) + i \cdot sin (ϕ_{1})]$ $z_1=r_1[cos(phi_1)+i*sin(phi_1)]$
$z_{2} = r_{2} [cos (ϕ_{2}) + i \cdot sin (ϕ_{2})]$ $z_2=r_2[cos(phi_2)+i*sin(phi_2)]$
Then their product is
$z_{1} \cdot z_{2} = r_{1} [cos (ϕ_{1}) + i \cdot sin (ϕ_{1})] \cdot r_{2} [cos (ϕ_{2}) + i \cdot sin (ϕ_{2})]$ $z_1*z_2=r_1[cos(phi_1)+i*sin(phi_1)]*r_2[cos(phi_2)+i*sin(phi_2)]$
Performing multiplication on the right, replacing $i^{2}$ $i^2$ with $- 1$ $-1$ and using trigonometric formulas for cosine and sine of a sum of two angles, we obtain
$z_{1} \cdot z_{2} = r_{1} r_{2} [cos (ϕ_{1} + ϕ_{2}) + i \cdot sin (ϕ_{1} + ϕ_{2})]$ $z_1*z_2=r_1r_2[cos(phi_1+phi_2)+i*sin(phi_1+phi_2)]$
The above is a polar representation of a product of two complex numbers represented in polar form.

Raising to any real power is also very convenient in polar form as this operation is an extension of multiplication:
${r [cos (ϕ) + i \cdot sin (ϕ)]}^{t} = r^{t} [cos (t \cdot ϕ) + i \cdot sin (t \cdot ϕ)]$ ${r[cos(phi)+i*sin(phi)]}^t=r^t[cos(t*phi)+i*sin(t*phi)]$

Addition of complex numbers is much more convenient in canonical form $z = a + i \cdot b$ $z=a+i*b$ . That's why, to add two complex numbers in polar form, we can convert polar to canonical, add and then convert the result back to polar form.
The first step (getting a sum in canonical form) results is
$z_{1} + z_{2} = [r_{1} cos (ϕ_{1}) + r_{2} cos (ϕ_{2})] + i \cdot [r_{1} sin (ϕ_{1}) + r_{2} sin (ϕ_{2})]$ $z_1+z_2=[r_1cos(phi_1)+r_2cos(phi_2)]+i*[r_1sin(phi_1)+r_2sin(phi_2)]$

Converting this to a polar form can be performed according to general rule of obtaining modulus (absolute value) and argument (phase) of a complex number represented as $z = a + i \cdot b$ $z=a+i*b$ where
$a = r_{1} cos (ϕ_{1}) + r_{2} cos (ϕ_{2})$ $a=r_1cos(phi_1)+r_2cos(phi_2)$ and
$b = r_{1} sin (ϕ_{1}) + r_{2} sin (ϕ_{2})$ $b=r_1sin(phi_1)+r_2sin(phi_2)$

This general rule states that
$z = r [cos (ϕ) + i \cdot sin (ϕ)]$ $z=r[cos(phi)+i*sin(phi)]$ where
$r = \sqrt{a^{2} + b^{2}}$ $r=sqrt(a^2+b^2)$ and
angle $ϕ$ $phi$ (usually, in radians) is defined by its trigonometric functions
$sin (ϕ) = \frac{b}{r}$ $sin(phi)=b/r$ ,
$cos (ϕ) = \frac{a}{r}$ $cos(phi)=a/r$
(it's not defined only if both $a = 0$ $a=0$ and $b = 0$ $b=0$ ).
Alternatively, we can use these equations to define angle $ϕ$ $phi$ :
If $a \neq 0$ $a!=0$ , $tan (ϕ) = \frac{b}{a}$ $tan(phi)=b/a$ . Or, if $b \neq 0$ $b!=0$ , $cot (ϕ) = \frac{a}{b}$ $cot(phi)=a/b$ .

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Is it possible to perform basic operations on complex numbers in polar form?

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