If z1=6+3 i and z 2=2- i, then ( z 1/ z 2) is equal to

Question

If z1=6+3 i and z 2=2- i, then ( z 1/ z 2) is equal to (A) (1/5)(9+12 i ) (B) 9+12 i (C) 3+2 i (D) (1/5)(12+9 i)

Tardigrade Admin · Accepted Answer

Let z1=6+3 i and z2=2-i Then,(z1/z2)=(6+3 i) (1/2-i)=((6+3 i)(2+i)/(2-i)(2+i)) =(6+3 i)((2/22+(-1)2)+i (1/22+(-1)2)) =(6+3 i)((2/5)+i (1/5))=(6+3 i) ((2+i)/5) =(1/5)[12-3+ i (6+6)]=(1/5)(9+12 i )

Q. If $z_{1} = 6 + 3 i$ and $z_{2} = 2 - i$ , then $\frac{z _{1}}{z _{2}}$ is equal to

$\frac{1}{5} (9 + 12 i)$

$9 + 12 i$

$3 + 2 i$

$\frac{1}{5} (12 + 9 i)$

Q. If z1​=6+3i and z2​=2−i, then z2​z1​​ is equal to

51​(9+12i)

9+12i

3+2i

51​(12+9i)

Let z1​=6+3i and z2​=2−i Then,z2​z1​​=(6+3i)2−i1​=(2−i)(2+i)(6+3i)(2+i)​ =(6+3i)(22+(−1)22​+i22+(−1)21​) =(6+3i)(52​+i51​)=(6+3i)5(2+i)​ =51​[12−3+i(6+6)]=51​(9+12i)

Q. If $z_{1} = 6 + 3 i$ and $z_{2} = 2 - i$ , then $\frac{z _{1}}{z _{2}}$ is equal to

$\frac{1}{5} (9 + 12 i)$

$9 + 12 i$

$3 + 2 i$

$\frac{1}{5} (12 + 9 i)$