If z1=1+2i and z2=3+5i, then Re[ barz2z1/z2] is equal to

Question

If z1=1+2i and z2=3+5i, then Re[ barz2z1/z2] is equal to (A)  -31/17  (B)  17/22  (C)  -17/31  (D)  22/17

Tardigrade Admin · Accepted Answer

We have, z1=1+2 i and z2=3+5 i ∴ barz2=3-5 i Now, barz2 z1=(3-5 i)(1+2 i) =3+6 i-5 i-10 i2 =3+i+10=13+i ∴ ( barz2 z1/z2)=((13+i)/(3+5 i)) × ((3-5 i)/(3-5 i)) =(39-65 i+3 i-5 i2/9+25) =(44-62 i/34) ∴ Real part of (( barz2 z1/z2))=(44/34)=(22/17)

Q. If $z_{1} = 1 + 2 i$ and $z_{2} = 3 + 5 i,$ then $R e [\overset{z}{ˉ}_{2} z_{1} / z_{2}]$ is equal to

$- 31/17$

$17/22$

$- 17/31$

$22/17$

Q. If z1​=1+2i and z2​=3+5i, then Re[zˉ2​z1​/z2​] is equal to

−31/17

17/22

−17/31

22/17

We have, z1​=1+2i and z2​=3+5i ∴zˉ2​=3−5i Now, zˉ2​z1​=(3−5i)(1+2i) =3+6i−5i−10i2 =3+i+10=13+i ∴z2​zˉ2​z1​​=(3+5i)(13+i)​×(3−5i)(3−5i)​ =9+2539−65i+3i−5i2​ =3444−62i​ ∴ Real part of (z2​zˉ2​z1​​)=3444​=1722​

Q. If $z_{1} = 1 + 2 i$ and $z_{2} = 3 + 5 i,$ then $R e [\overset{z}{ˉ}_{2} z_{1} / z_{2}]$ is equal to

$- 31/17$

$17/22$

$- 17/31$

$22/17$