If z_1=1-2i ;z_2=1+i and z_3=3+4i, then ( 1z_1+ 3z_2) z_3z_2=

Question

Accepted Answer

Correct answer: D. (d) Given that, z_1=1-2i ,z_2=1+i ,z_3=3+4i Now, ( 1z_1+ 3z_2) z_3z_2=( 11-2i+ 31+i) ( 3+4i1+i ) = (1+i )+3 (1-2i ) (1-2i ) (1+i ) 3+4i1+i= 1+i+3-6i1+i-2i-2i^2 3+4i1+i = 4-5i3-i 3+4i1+i = 12+16i-15i+203+3i-i+1= 32+i4+2i= 32+i2 (2+i ) = (32+i)2(2+i) 2-i2-i= 64+2i-32i+12(4+1) = 65-30i10= 132-3i

$If z_{1} = 1 - 2 i; z_{2} = 1 + i and z_{3} = 3 + 4 i, then (\frac{1}{z _{1}} + \frac{3}{z _{2}}) \frac{z _{3}}{z _{2}} =$

Solution

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