Answer:
[tex]\frac{1}{3\sqrt{2} } (1+i)[/tex]
Step-by-step explanation:
Given are two complex numbers as
z1 = 3 (cos 135+isin 135)
z2 = 9(cos 45+isin45)
To find quotient
We can use Demoivre theorem for products and quotients here
[tex]\frac{z1}{z2} =\frac{3(cos135+isin135)}{9(cos45+isin45)} \\=\frac{1}{3} (cos135-90+isin 135-90)\\=\frac{1}{3} (cos45 +isin 45)\\=\frac{1}{3\sqrt{2} } (1+i)[/tex]
Answer:
Quotient is i/3.
Step-by-step explanation:
Given:
Complex numbers are 3( cos 135° + i sin 135° ) and 9( cos 45° + i sin 45° )
To find: Quotient of the given complex number.
Consider,
[tex]\frac{3(cos\,135+i\:sin\,135)}{9(cos\,45+i\:sin\,45)}[/tex]
[tex]\frac{3}{9}\times\frac{cos\,135+i\:sin\,135}{cos\,45+i\:sin\,45}[/tex]
[tex]=\frac{1}{3}\times\frac{cos\,135+i\:sin\,135}{cos\,45+i\:sin\,45}\times\frac{cos\,45-i\:sin\,45}{cos\,45-\:sin\,45)}[/tex]
[tex]=\frac{1}{3}\times\frac{(cos\,135+i\:sin\,135)(cos\,45-i\:sin\,45)}{(cos\,45-i\:sin\,45)(cos\,45+\:sin\,45)}[/tex]
[tex]=\frac{1}{3}\times\frac{(cos\,135\:cos\,45+sin\,135\:sin\,45+i(sin\,135\:cos\,45-cos\,135\:\:sin\,45)}{cos^2\,45-(i\:sin45)^2}[/tex]
using, cos A cos B + sin A sin B = cos( A - B ) and sin A cos B - cos A sin B = sin( A - B )
[tex]=\frac{1}{3}\times\frac{cos\,(135-45)+i\:sin\,(135-45)}{cos^2\,45-(-1)sin^2\,45}[/tex]
[tex]=\frac{1}{3}\times\frac{cos\,90+i\:sin\,90}{cos^2\,45+sin^2\,45}[/tex]
[tex]=\frac{1}{3}\times\frac{0+i}{1}[/tex]
[tex]=\frac{i}{3}[/tex]
Therefore, Quotient is i/3.
