Let z 1-2- i - z 2-2- i- Findi Re z 1 z 2-z1ii Im1-z1z1

Here z_1=2 i, z_2= 2+i ∴ z̅_1=2+i (i) Now z_1z_2=(2 i)( 2+i)= 4+2i+2i i^2 =( 4+1)+4i= 3+4i ∴ =z_1z_2/z̅_̅1̅= 3+4i/2+i×2 i/2 i = 6+3i+8i 4i^2/4 i^2 =( 6+4)+11i ...

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

Byju's Answer

Standard XI

Mathematics

Properties of Conjugate of a Complex Number

Let z 1=2- i ...

Question

Let $z_{1} = 2 - i, z_{2} = - 2 + i$ . Find

$(i) R e (\frac{z_{1} z_{2}}{_{1}})$

$(i i) I m (\frac{1}{z_{1}_{1}})$

Open in App

Solution

Here $z_{1} = 2 - i, z_{2} = - 2 + i ∴ {¯ z}_{1} = 2 + i$

(i) Now $z_{1} z_{2} = (2 - i) (- 2 + i) = - 4 + 2 i + 2 i - i^{2} = (- 4 + 1) + 4 i = - 3 + 4 i$

$∴ = \frac{z_{1} z_{2}}{_{1}} = \frac{- 3 + 4 i}{2 + i} \times \frac{2 - i}{2 - i}$

$= \frac{- 6 + 3 i + 8 i - 4 i^{2}}{4 - i^{2}}$

$= \frac{(- 6 + 4) + 11 i}{4 + 1} = \frac{- 2 + 11 i}{5}$

$= \frac{- 2}{5} + \frac{11}{5} i$

$∴ R e (\frac{z_{1} z_{2}}{_{1}}) = \frac{- 2}{5} .$

$(i i) \frac{1}{z_{1}_{1}} = \frac{1}{(2 - i) (2 + i)} = \frac{1}{4 - i^{2}} = \frac{1}{5}$

$∴ I m (\frac{1}{z_{1}_{1}}) = 0$

Suggest Corrections

Similar questions

$I f z_{1} = 2 - i, z_{2} = - 2 + i, find (i) R e (\frac{z_{1} z_{2}}{z_{1}}) (i i) I m (\frac{1}{z_{1} z_{2}})$

Q. If

z_{1} = 2 - i, z_{2} = - 2 + i

,find

R e (\frac{z_{1} z_{2}}{_{1}})

Q. Let

z_{1}, z_{2} \in C

and

x = | z_{1} z_{2} | - R e (z_{1} z_{2}) - \frac{1}{2} |_{2} - z_{1} |^{2} + \frac{1}{2} (| z_{2} | - | z_{1} |)^{2}

then

Q. If

z_{1} = 2 - i, z_{2} = - 2 + i

,find

I m (\frac{1}{z_{1}_{2}})

Q. If

z_{1} = 2 - i, z_{2} = - 2 + i,

find
(i) Re

(\frac{z_{1} z_{2}}{z_{1}})

(ii) Im

(\frac{1}{z_{1} z_{1}})

Join BYJU'S Learning Program

Let z1=2−i,z2=−2+i. Find (i) Re(z1z2¯z1) (ii) Im (1z1¯z1)

Here z1=2−i,z2=−2+i∴ ¯z1=2+i (i) Now z1z2=(2−i)(−2+i)=−4+2i+2i−i2=(−4+1)+4i=−3+4i ∴ =z1z2¯z1=−3+4i2+i×2−i2−i =−6+3i+8i−4i24−i2 =(−6+4)+11i4+1=−2+11i5 =−25+115i ∴ Re(z1z2¯z1)=−25. (ii) 1z1¯z1=1(2−i)(2+i)=14−i2=15 ∴ Im(1z1¯z1)=0

Let $z_{1} = 2 - i, z_{2} = - 2 + i$ . Find

$(i) R e (\frac{z_{1} z_{2}}{_{1}})$

$(i i) I m (\frac{1}{z_{1}_{1}})$