Assertion A - z1-1-i- z2-1-i then the polar form of z1-z2 is 1-2Reason R - z1-r1-1- z2-r2-2 then z1-z2 is represented as a point in the polar form r1-r2-1-2

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Sum of Coefficients of All Terms

Assertion A :...

Question

Assertion (A) : $z_{1} = 1 + i, z_{2} = 1 - i$ then the polar form of $\frac{z_{1}}{z_{2}}$ is $(1, \frac{π}{2})$
Reason (R) : $z_{1} = (r_{1}, θ_{1}), z_{2} = (r_{2}, θ_{2})$ then $\frac{z_{1}}{z_{2}}$ is represented as a point in the polar form $(\frac{r_{1}}{r_{2}}, θ_{1} - θ_{2})$

Both A and R are true and R is the correct explanation of A

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Both A and R are true but R is not the correct explanation of A

No worries! We‘ve got your back. Try BYJU‘S free classes today!

A is false, R is true

No worries! We‘ve got your back. Try BYJU‘S free classes today!

A is true R is false

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

Suggest Corrections

Similar questions

z_{1}, z_{2}, z_{3}

\frac{p}{| z_{2} - z_{3} |} = \frac{q}{| z_{3} - z_{1} |} = \frac{r}{| z_{1} - z_{2} |}

\frac{p^{2}}{z_{2} - z_{3}} + \frac{q^{2}}{z_{3} - z_{1}} + \frac{r^{2}}{z_{1} - z_{2}} = 0

z_{1} = 1 - i

z_{2} = \sqrt{3} + i

z_{1} \cdot z_{2}

z_{1}, z_{2}, z_{3}

p, q, r

\frac{p}{| z_{2} - z_{3} |} = \frac{q}{| z_{3} - z_{1} |} = \frac{r}{| z_{1} - z_{2} |}

\frac{p^{2}}{z_{2} - z_{3}} + \frac{q^{2}}{z_{3} - z_{1}} + \frac{r^{2}}{z_{1} - z_{2}} = 0

z_{1}, z_{2}

∣ ∣ ∣ \frac{z_{1} - z_{2}}{z_{1} + z_{2}} ∣ ∣ ∣ = 1

t z_{1} = k z_{2}

k \in R

(z_{1} - z_{2})

(z_{1} + z_{2})

z_{1} & z_{2}

a r g (\frac{z_{1} + z_{2}}{z_{1} - z_{2}}) = \frac{π}{2}

| z_{1} + z_{2} | \neq | z_{1} - z_{2} |

z_{1}, z_{2}, z_{1} + z_{2}

Join BYJU'S Learning Program