Convert the complex number 41−√3i into polar form.
Answer:
2(cosπ3+i sinπ3)
- 41−√3i=41−√3i×1+√3i1+√3i[ Rationalizing the denominator ]=4(1+√3i)(1)2−(√3i)2[ Since (a+b)(a−b)=a2−b2]=4(1+√3i)1+3[ Since i2=−1]=4(1+√3i)4=1+√3i
- Let, z=1+√3i
The standard polar form of a complex number is r(cosθ+i sinθ)
- On comparing z with the standard polar form of a complex number, we get,
r cos θ=1 and r sin θ=√3
Now, r cos θ=1…(1)⟹r2 cos2θ=12…(2)r sinθ=√3…(3)⟹r2 sin2θ=√32…(4) On Adding (2) and (4) we get,
r2 cos2θ+r2 sin2θ=12+√32⟹r2(cos2θ+sin2θ)=1+3⟹r2=4[Since, cos2θ+sin2θ=1] ⟹r=2[Conventionally r>0] - Substituting the value of r in eq (1) and (3) we get,
cosθ=12 and sinθ=√32
⟹θ=π3 - Hence, the polar form of the complex number z=1+√3i is 2(cosπ3+i sinπ3).