z1=[cos pi/4 +sin pi/4] z2= [cos 2pi+sin 2pi]

z1= sqrt 5 solve.

how do you do this? substitute z1 and do what?

1 answer

actually, z1 = cos pi/4 +sin pi/4 i
that is (1,π/4)
z2 = (1,2π) = (1,0)

no idea how you can say z1=√5, when you already said is is (1,π/4)

z1*z2 = (1,π/4)
z1+z2 = (1/√2 + 1/√2 i)+(1+0i) = (1+1/√2)+1/√2 i

|z1+z2| = √((1+1/√2)^2 + 1/2) = √(2+√2)

I can't fathom just what the question is, much less the answer...
Similar Questions
  1. When I solve the inquality 2x^2 - 6 < 0,I get x < + or - sqrt(3) So how do I write the solution? Is it (+sqrt(3),-sqrt(3)) or
    1. answers icon 0 answers
    1. answers icon 2 answers
  2. Solve in the exact form.(sqrt of 4x+1)+(sqrt of x+1)=2 Someone showed me to do this next: Square both sides..so.. 4x+1+2((sqrt
    1. answers icon 4 answers
  3. Could someone show me how to solve these problems step by step....I am confused on how to fully break this down to simpliest
    1. answers icon 1 answer
more similar questions