Modulus and argument of the
complex number
number z=1+i/1-i is
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Modulus and argument of the
complex number
number z=1+i/1-i is
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Given ,
The complex number is
On simplifying , we get
[tex] \tt \implies \frac{(1 + i)(1 + i)}{(1 - i)(1 + i)} [/tex]
[tex] \tt \implies\frac{ {(1 + i)}^{2} }{ {(1)}^{2} - {(i)}^{2} } [/tex]
[tex] \tt \implies\frac{ {(1)}^{2} + {(i)}^{2} + 2(1)(i) }{1 - ( - 1)} [/tex]
[tex] \tt \implies \frac{1 + ( - 1) + 2i}{2} [/tex]
[tex] \tt \implies\frac{2i}{2} [/tex]
[tex]\tt \implies i[/tex]
The given complex number in standard form is (0 + i)
On comparing with rCosΦ + irSinΦ) , we get
rCosΦ = 0 --- (i)
rSinΦ = 1 --- (ii)
Squaring eq (i) and eq (ii) , we get
(r)²Cos²(Φ) = 0 --- (iii)
(r)²Sin²(Φ) = (1)² --- (iv)
Adding eq (iii) and eq (iv) , we get
1 = (r)²{Cos²(Φ) + Sin²(Φ)}
1 = (r)²{1}
(r)² = 1
r = 1 , because r > 0
Hence , the modulus of complex number is 1
Put the value of r = 1 in eq (i) , we get
(1)CosΦ = 0
CosΦ = 0
CosΦ = Cos(90)
Φ = 90
Hence , the argument or amplitude of complex number is 1 and 90°