z/2+z/3+z/4 =13 ,zis equal to
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Answer :
Given: z = 2 – 3i
Given: z = 2 – 3iTo Prove: z2 – 4z + 13 = 0
Given: z = 2 – 3iTo Prove: z2 – 4z + 13 = 0Taking LHS, z2 – 4z + 13
Given: z = 2 – 3iTo Prove: z2 – 4z + 13 = 0Taking LHS, z2 – 4z + 13Putting the value of z = 2 – 3i, we get
Given: z = 2 – 3iTo Prove: z2 – 4z + 13 = 0Taking LHS, z2 – 4z + 13Putting the value of z = 2 – 3i, we get(2 – 3i)2 – 4(2 – 3i) + 13
Given: z = 2 – 3iTo Prove: z2 – 4z + 13 = 0Taking LHS, z2 – 4z + 13Putting the value of z = 2 – 3i, we get(2 – 3i)2 – 4(2 – 3i) + 13= 4 + 9i2 – 12i – 8 + 12i + 13
Given: z = 2 – 3iTo Prove: z2 – 4z + 13 = 0Taking LHS, z2 – 4z + 13Putting the value of z = 2 – 3i, we get(2 – 3i)2 – 4(2 – 3i) + 13= 4 + 9i2 – 12i – 8 + 12i + 13= 9(-1) + 9
Given: z = 2 – 3iTo Prove: z2 – 4z + 13 = 0Taking LHS, z2 – 4z + 13Putting the value of z = 2 – 3i, we get(2 – 3i)2 – 4(2 – 3i) + 13= 4 + 9i2 – 12i – 8 + 12i + 13= 9(-1) + 9= - 9 + 9
Given: z = 2 – 3iTo Prove: z2 – 4z + 13 = 0Taking LHS, z2 – 4z + 13Putting the value of z = 2 – 3i, we get(2 – 3i)2 – 4(2 – 3i) + 13= 4 + 9i2 – 12i – 8 + 12i + 13= 9(-1) + 9= - 9 + 9= 0
Given: z = 2 – 3iTo Prove: z2 – 4z + 13 = 0Taking LHS, z2 – 4z + 13Putting the value of z = 2 – 3i, we get(2 – 3i)2 – 4(2 – 3i) + 13= 4 + 9i2 – 12i – 8 + 12i + 13= 9(-1) + 9= - 9 + 9= 0= RHS
Given: z = 2 – 3iTo Prove: z2 – 4z + 13 = 0Taking LHS, z2 – 4z + 13Putting the value of z = 2 – 3i, we get(2 – 3i)2 – 4(2 – 3i) + 13= 4 + 9i2 – 12i – 8 + 12i + 13= 9(-1) + 9= - 9 + 9= 0= RHS= 4z3 – 16z2 + 52z + 13z2 – 52z + 169
Given: z = 2 – 3iTo Prove: z2 – 4z + 13 = 0Taking LHS, z2 – 4z + 13Putting the value of z = 2 – 3i, we get(2 – 3i)2 – 4(2 – 3i) + 13= 4 + 9i2 – 12i – 8 + 12i + 13= 9(-1) + 9= - 9 + 9= 0= RHS= 4z3 – 16z2 + 52z + 13z2 – 52z + 169= 4z(z2 – 4z + 13) + 13 (z2 – 4z + 13)
Given: z = 2 – 3iTo Prove: z2 – 4z + 13 = 0Taking LHS, z2 – 4z + 13Putting the value of z = 2 – 3i, we get(2 – 3i)2 – 4(2 – 3i) + 13= 4 + 9i2 – 12i – 8 + 12i + 13= 9(-1) + 9= - 9 + 9= 0= RHS= 4z3 – 16z2 + 52z + 13z2 – 52z + 169= 4z(z2 – 4z + 13) + 13 (z2 – 4z + 13)= 4z(0) + 13(0) [from eq. (i)]
Given: z = 2 – 3iTo Prove: z2 – 4z + 13 = 0Taking LHS, z2 – 4z + 13Putting the value of z = 2 – 3i, we get(2 – 3i)2 – 4(2 – 3i) + 13= 4 + 9i2 – 12i – 8 + 12i + 13= 9(-1) + 9= - 9 + 9= 0= RHS= 4z3 – 16z2 + 52z + 13z2 – 52z + 169= 4z(z2 – 4z + 13) + 13 (z2 – 4z + 13)= 4z(0) + 13(0) [from eq. (i)]= 0
Given: z = 2 – 3iTo Prove: z2 – 4z + 13 = 0Taking LHS, z2 – 4z + 13Putting the value of z = 2 – 3i, we get(2 – 3i)2 – 4(2 – 3i) + 13= 4 + 9i2 – 12i – 8 + 12i + 13= 9(-1) + 9= - 9 + 9= 0= RHS= 4z3 – 16z2 + 52z + 13z2 – 52z + 169= 4z(z2 – 4z + 13) + 13 (z2 – 4z + 13)= 4z(0) + 13(0) [from eq. (i)]= 0= RHS
Given: z = 2 – 3iTo Prove: z2 – 4z + 13 = 0Taking LHS, z2 – 4z + 13Putting the value of z = 2 – 3i, we get(2 – 3i)2 – 4(2 – 3i) + 13= 4 + 9i2 – 12i – 8 + 12i + 13= 9(-1) + 9= - 9 + 9= 0= RHS= 4z3 – 16z2 + 52z + 13z2 – 52z + 169= 4z(z2 – 4z + 13) + 13 (z2 – 4z + 13)= 4z(0) + 13(0) [from eq. (i)]= 0= RHSHence Proved
Answer:
Z = 12
Step-by-step explanation:
Let's solve your equation step-by-step.
z /2 + z /3 + z /4 =13
Step 1: Simplify both sides of the equation.
z /2 + z /3 + z /4 =13
1 /2 z+ 1 /3 z+ 1 /4 z=13
( 1 /2 z+ 1 /3 z+ 1 /4 z)=13(Combine Like Terms)
13 /12 z=13
Step 2: Multiply both sides by 12/13.
( 12 /13 )*( 13 /12 z)=( 12 /13 )*(13)
z=12