Question

Which of the following is/are true, if z, z₁, z₂ are complex numbers?

• A.
• B.
• C.
• D. All of these

Answer 1

Answer: (A), (B)

The correct options are
A

B

Let z = r ($\cos \theta +i\sin \theta$)

$\overline{Z}$ = r ($\cos \theta -i\sin \theta$)

Option A: ($\overline{\overline{Z}}$) = r ($\cos \theta +i\sin \theta$) = z

Option B: $z^n$ = $r^n$ ${(\cos \theta +i\sin \theta )}^n$

$z^n$ = r ($\cos \theta +i\sin \theta$) {Use de moivre's theorem ${(\cos \theta +i\sin \theta )}^n$ = ($\cos \theta +i\sin \theta$) }

($\overline{Z^n}$) = $r^n$ ($\cos n\theta +i\sin n\theta$)

($\overline{Z}{)}^n$ = $r^n$ ${(\cos \theta -i\sin \theta )}^n$

($\overline{Z}{)}^n$ = $r^n$ ($\cos n\theta +i\sin n\theta$) { Use de moivre's theorem ${(\cos \theta +i\sin \theta )}^n$ = ($\cos \theta -i\sin \theta$) }

So, ($\overline{Z^n}$) = ($\overline{Z}{)}^n$

Let $z_1$ = $r_1$ ($\cos {\theta }_1+i\sin {\theta }_1$) = $r_1$ $e^{i{\theta }_1}$

$z_2$ = $r_2$ ($\cos {\theta }_2+i\sin {\theta }_2$) = $r_2$ $e^{i{\theta }_2}$

$z_1.z_2$ = $r_1$.$r_2$ [($\cos ({\theta }_1+{\theta }_2)+i\sin ({\theta }_1+{\theta }_2)$)]

($\overline{z_1.z_2}$) = $r_1$.$r_2$ [($\cos ({\theta }_1+{\theta }_2)-i\sin ({\theta }_1+{\theta }_2)$)]

So, We see that

($\overline{z_1.z_2}$) $\ne$ $z_1.z_2$

Only option A and B are correct.