Question

) If the polar form of z = 8(cosπ/3+i sinπ/3) then the Cartesian form is *​

Answer 1

Answer:

Solution :

z=i−1cos(π3)+isin(π3)∗cos(π3)−isin(π3)cos(π3)−isin(π3)

=icos(π3)−i2(sin(π3))−cos(π3)+isin(π3)cos2(π3)−i2sin2(π3)

=sin(π3)−cos(π3)+i(sin(π3)+cos(π3))14+34

⇒z=sin(π3)−cos(π3)+i(sin(π3)+cos(π3))

∴rcosθ=sin(π3)−cos(π3)

rsinθ=sin(π3)+cos(π3)

∴r2(cos2θ+sin2θ)=(sin(π3)−cos(π3))2+(sin(π3)+cos(π3))2

⇒r2=2(sin2(π3)+cos2(π3))=2

⇒r=2–√

∴2–√cosθ=sin(π3)−cos(π3)=3–√2−12

⇒cosθ=3–√−122–√=cos(5π12)

⇒θ=5π12

⇒2–√sinθ=sin(π3)+cos(π3)=3–√2+12

⇒sinθ=3–√+122–√=sin(5π12)

⇒θ=sin(5π12)

∴z=2–√(cos(5π12)+isin(5π12)), which is the required polar form.

Step-by-step explanation:

Answer 2

Answer:

z=i−1cos(π3)+isin(π3)∗cos(π3)−isin(π3)cos(π3)−isin(π3)

=icos(π3)−i2(sin(π3))−cos(π3)+isin(π3)cos2(π3)−i2sin2(π3)

=sin(π3)−cos(π3)+i(sin(π3)+cos(π3))14+34

⇒z=sin(π3)−cos(π3)+i(sin(π3)+cos(π3))

∴rcosθ=sin(π3)−cos(π3)

rsinθ=sin(π3)+cos(π3)

∴r2(cos2θ+sin2θ)=(sin(π3)−cos(π3))2+(sin(π3)+cos(π3))2

⇒r2=2(sin2(π3)+cos2(π3))=2

⇒r=2–√

∴2–√cosθ=sin(π3)−cos(π3)=3–√2−12

⇒cosθ=3–√−122–√=cos(5π12)

⇒θ=5π12

⇒2–√sinθ=sin(π3)+cos(π3)=3–√2+12

⇒sinθ=3–√+122–√=sin(5π12)

⇒θ=sin(5π12)

∴z=2–√(cos(5π12)+isin(5π12)), which is the required polar form.