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Properties that Modulus of a complicated NumberLet z be any complicated number, then(I) |-z| = |z |Example : Let z = 7 + 8i-z = -( 7 + 8i)-z = -7 -8imodulus the (-z) =|-z| =$\sqrt(-7)^2 + (-8)^2$=$\sqrt49 + 64$ =$\sqrt113$modulus that (z) = |z|=$\sqrt7^2 + 8^2$=$\sqrt49 + 64$ =$\sqrt113$So indigenous the over we have the right to say that |-z| = |z |(II) |z| = 0 if, z = 0Proof : If z = a+ib ⇒ |z| = $\sqrta^2 + b^2$|z| = 0 ⇒ $\sqrta^2 + b^2$ = 0 ⇒ $a^2 + b^2$ = 0So, $a^2$ = 0 and also $b^2$ = 0 ⇒ a = 0 and also b = 0i.e. Z = 0 + i0 = 0So |z| = 0 if, z = 0(III) The pure of a product that two complicated numbers z1 and z2 is same to the product that the absolute worths of the numbers. I.e$\left |z1.z2 \right | $= $\left | z1 \right . | $ $\left | z2 \right | $(IV) The absolute of a quotient the two complicated numbers z1 and z2 (≠ 0) is equal to the quotient that the absolute values of the dividend and the divisor.$\left | \fracz1z2 \right |$= $\frac \left $(V) The pure of the amount of two conjugate complex numbers z1 and z2 deserve to never exceed the amount of their absolute values, i.e.$\left | z1+z2 \right |$ $\leq$ $ \left | z1 \right |$ +$\left |z2 \right | $This inequality is referred to as triangle inequality.11th grade mathFrom modulus of a facility number to Home
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