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Property 3

arg(z1*z2) = ...

Question

arg(z1*z2) = arg(z1) + arg(z2)

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Solution

$a r g (z_{1} . z_{2}) = a r g (z_{1}) + a r g (z_{2}) L e t z_{1} = r_{1} (\cos θ + i \sin θ) a n d z_{2} = r_{2} (\cos α + i \sin α) A n d |z_{1}| = r_{1}, |z_{2}| = r_{2}, a n d a r g (z_{1}) = θ, a r g (z_{2}) = α S o z_{1} . z_{2} = r_{1} (\cos θ + i \sin θ) \times r_{2} (\cos α + i \sin α) = r_{1} r_{2} (\cos θ \cos α - \sin θ \sin α) + i (\sin θ \cos α + \sin α \cos θ) = r_{1} r_{2} [\cos (α + θ) + i \sin (α + θ)] H e n c e a r g (z_{1} . z_{2}) = a r g (z_{1}) + a r g (z_{2})$

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Similar questions

a r g z_{1} = \frac{π}{2}

a r g {¯ ¯ ¯ z}_{2} = \frac{π}{4}

a r g (\frac{z_{1}}{z_{2}}) =

z_{1}, z_{2}

are complex numbers then the correct match List- I to List II is:

List - I	List - II
A) arg z_1z_2	1) $a r g z_{1} - a r g z_{2}$
B) $a r g z_{1}_{2}$	2) $a r g z_{1} - a r g z_{2} = \frac{π}{2}$
C) $\| z_{1} + z_{2} \| = \| z_{1} - z_{2} \|$	3) $a r g z_{1} = a r g z_{2}$
D) $\| z_{1} + z_{2} \|^{2}$	4) $a r g z_{1} + a r g z_{2}$
E) $\| z_{1} + z_{2} \| = \| z_{1} \| + \| z_{2} \|$	5) $r_{1}^{2} + r_{2}^{2} + 2 r_{1} r_{2} c o s (θ_{1} - θ_{2})$

| z_{1} + z_{2} | = | z_{1} | + | z_{2} |

a r g z_{1} - a r g z_{2} =

| z_{1} + z_{2} | = | z_{1} | + | z_{2} |

, then show that

a r g (z_{1}) = a r g (z_{2})

| z_{1} - z_{2} | = | z_{1} | + | z_{2} |

, then prove that

a r g (z_{1}) - a r g (z_{2}) = π

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