Let z=((2 √3+2 i)8/(1-i)6)+((1+i)6/(2 √3-2 i)8). Then argument of z is

Question

Let z=((2 √3+2 i)8/(1-i)6)+((1+i)6/(2 √3-2 i)8). Then argument of z is (A) (5 π/6) (B) (π/6) (C) -(π/6) (D) -(5 π/6)

Tardigrade Admin · Accepted Answer

z=(z18/ barz26)+(z26/ barz18)=(|z1|16+|z2|12/( barz2)6( barz1)8) ∴ arg (z)=-8 arg ( barz1)-6 arg ( barz2)+2 k π, k ∈ I =8 arg (z1)+6 arg (z2)+2 k π =8 ⋅ (π/6)+6 ⋅ (π/4)+2 k π =(4 π/3)+(3 π/2)-2 π=(5 π/6)

Q. Let $z = \frac{( 2 3 + 2 i ) ^{8}}{( 1 - i ) ^{6}} + \frac{( 1 + i ) ^{6}}{( 2 3 - 2 i ) ^{8}}$ . Then argument of $z$ is

$\frac{5 π}{6}$

$\frac{π}{6}$

$- \frac{π}{6}$

$- \frac{5 π}{6}$

Q. Let z=(1−i)6(23​+2i)8​+(23​−2i)8(1+i)6​. Then argument of z is

65π​

6π​

−6π​

−65π​

z=zˉ26​z18​​+zˉ18​z26​​=(zˉ2​)6(zˉ1​)8∣z1​∣16+∣z2​∣12​ ∴arg(z)=−8arg(zˉ1​)−6arg(zˉ2​)+2kπ,k∈I =8arg(z1​)+6arg(z2​)+2kπ =8⋅6π​+6⋅4π​+2kπ =34π​+23π​−2π=65π​

Q. Let $z = \frac{( 2 3 + 2 i ) ^{8}}{( 1 - i ) ^{6}} + \frac{( 1 + i ) ^{6}}{( 2 3 - 2 i ) ^{8}}$ . Then argument of $z$ is

$\frac{5 π}{6}$

$\frac{π}{6}$

$- \frac{π}{6}$

$- \frac{5 π}{6}$