Convert the complex number z- i-1 cos- 3 -i sin- 3 in the polar from and hence find the modulus and the argument of x-

i 1 can be written as √(2)(i/√(2) #160; #160; 1/√(2)) = √(2) (cos3π/4 isin3π/4 ) #160; = #160; √(2) e^i3π/4 #160; #160; ...

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Standard XII

Mathematics

Modulus of a Complex Number

Convert the c...

Question

Convert the complex number $z = \frac{i - 1}{cos \frac{π}{3} + i sin \frac{π}{3}}$ in the polar from and hence find the modulus and the argument of $x$ .

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Solution

$i$ - 1 can be written as $\sqrt{2} (i / \sqrt{2} - 1 / \sqrt{2})$

= $\sqrt{2} (c o s 3 π / 4 - i s i n 3 π / 4$ ) = $\sqrt{2} e^{i 3 π / 4}$ -(i)

$c o s π / 3 + i s i n π / 3$ in polar form can be written as $e^{i π / 3}$ -(ii)

Combining (i) and (ii),

z = $\sqrt{2} (e^{\frac{5 π}{12}})$

$\Rightarrow$ Modulus of z , |z| = $\sqrt{2}$
and Principal argument of z = $\frac{5 π}{12}$

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Convert the complex number z=i−1cosπ3+isinπ3 in the polar from and hence find the modulus and the argument of x.

i - 1 can be written as √2(i/√2−1/√2)=√2(cos3π/4−isin3π/4) = √2ei3π/4 -(i)cosπ/3+isinπ/3 in polar form can be written as eiπ/3 -(ii)Combining (i) and (ii),z = √2(e5π12)⇒ Modulus of z , |z| = √2and Principal argument of z = 5π12

Convert the complex number $z = \frac{i - 1}{cos \frac{π}{3} + i sin \frac{π}{3}}$ in the polar from and hence find the modulus and the argument of $x$ .