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Modulus of a Complex Number

For any two n...

Question

For any two non-zero complex numbers $z_{1}, z_{2}$ prove the inequality
$(| z_{1} | + | z_{2} |) ∣ ∣ ∣ \frac{z_{1}}{| z_{1} |} + \frac{z_{2}}{| z_{2} |} ∣ ∣ ∣ \leq 2 (| z_{1} | + | z_{2} |)$

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Solution

we have
$(| z_{1} | + | z_{2} |) ∣ ∣ ∣ \frac{z_{1}}{| z_{1} |} + \frac{z_{2}}{| z_{2} |} ∣ ∣ ∣ \leq (| z_{1} | + | z_{2} |) [∣ ∣ ∣ \frac{z_{1}}{| z_{1} |} + \frac{z_{2}}{| z_{2} |} ∣ ∣ ∣]$
$= (| z_{1} | + | z_{2} |) ∣ ∣ ∣ \frac{| z_{1} |}{| z_{1} |} + \frac{| z_{2} |}{| z_{2} |} ∣ ∣ ∣$
$= 2 (| z_{1} | + | z_{2} |)$ .

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Similar questions

Q. For any two non-zero complex numbers

Z_{1}

Z_{2}

(| Z_{1} | + | Z_{2} |) ∣ ∣ ∣ \frac{Z_{1}}{| Z_{1} |} + \frac{Z_{2}}{| Z_{2} |} ∣ ∣ ∣

Q. (a) For any two non-zero complex numbers

z_{1}

z_{2}

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

, then prove that

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

is zero.
(b) Prove the above result if we have

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

z_{1}

z_{2}

are two non-zero complex numbers such that

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

arg (z_{1}) - arg (z_{2}) =

z_{1}

z_{2}

are two complex numbers, then prove that

| z_{1} | + | z_{2} | = ∣ ∣ ∣ \frac{z_{1} + z_{2}}{2} + \sqrt{z_{1} z_{2}} ∣ ∣ ∣ + ∣ ∣ ∣ \frac{z_{1} + z_{2}}{2} - \sqrt{z_{1} z_{2}} ∣ ∣ ∣

z_{1}, z_{2}

be two non zero complex numbers satisfying the equation

∣ ∣ ∣ \frac{z_{1} + z_{2}}{z_{1} - z_{2}} ∣ ∣ ∣ = 1

\frac{z_{1}}{z_{2}} + (\frac{z_{1}}{z_{2}})

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Modulus of a Complex Number

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