Question

find the modulus and argument if Z = -2i​

Answer 1

$\mathtt{ \huge{ \fbox{SOLUTION :}}}$

Given ,

Complex number (z) = -2i

Real part of complex number (a) = 0

Imaginary part of complex number (b) = -2

We know that , the modulus (|z|) of complex number is given by

$\large \sf \fbox{ |z| = \sqrt{ {(a)}^{2} + {(b)}^{2} } }$

Substitute the known values , we get

$\sf \mapsto |z| = \sqrt{ {(0)}^{2} + {( - 2)}^{2} } \\ \\ \sf \mapsto |z| = \sqrt{4} \\ \\ \sf \mapsto |z| = 2$

Hence , the modulus of complex number is 2

We know that ,

$\sf \large \fbox{z = r( \cos \theta + i\sin \theta)}$

Thus ,

0 = rCos(Φ) ------- (i)

-2 = rSin(Φ) ------- (ii)

Thus , from equation (i) and (ii)

$\sf \mapsto 0 + (- 2 )= r\cos \theta + r \sin \theta \\ \\ \sf \: \: \: \: \: \: \: \: \: \: \: \: squaring \: both \: sides \:, \: we \: get \\ \\ \sf \mapsto {(0)}^{2} + {( - 2)}^{2} = {(r)}^{2} { \cos}^{2} \theta + {(r)}^{2} { \sin}^{2}\theta \\ \\ \sf \mapsto 4 = {(r)}^{2} ( { \cos}^{2} \theta + { \sin}^{2}\theta ) \\ \\ \sf \mapsto {(r)}^{2} = 4 \\ \\ \sf \mapsto r = \sqrt{4 } \\ \\ \sf \mapsto r = 2 \: \: \: \: \: \: \: \{ \because r > 0\}$

Put the value of r = 2 in equation (i) and (ii) , we get

$\sf \mapsto 0 = 2 \: \cos \theta \\ \\ \sf \mapsto \cos \theta = 0 \\ \\ \sf \mapsto \cos \theta =\cos (π + π/2 ) \\ \\ \sf \mapsto \theta = 3π/2$

Hence , the argument or amplitude of complex number is 3π/2