COMPLEX NUMBERS.. really need help !?

Question

I need to solve for the complex number z
the question is

(i+z)-3i(2-z)=iz+1

Ne help would be appreciated !! the answer is z= 11/5 + 3/5i.... i just need to know how to do it !

Thanx

Answer 1

Hmmm

Your aswer is correct. Below is the work
expanding we have:
i+z -6i+3iz=iz+1
then
z(1+2i)=1+5i
z=(1+5i)/(1+2i) (multiplying denominator and numerator by (1-2i)[complex conjugate]

we have
z=(11-3i)/(1+4)=
z=-11/5+(3/5)i

Answer 2

The sq. root of an imaginary variety is yet another imaginary variety. First be conscious that (a+bi)^2 = a^2 +2abi -b^2 = a^2-b^2 + (2ab)i So as an celebration, (2+i)^2 = 4 +4i -a million) =3+4i for this reason the sqrt of three+4i is two+i and a couple of-a million for the reason that complicated numbers continually are available conjugate pairs. So enable's see the thanks to reveal that the sqrt of three+4i is unquestionably 2 + i and a couple of - i shop in ideas (a+bi)^2= a^2 -b^2 + (2ab)i So we ought to instruct that a^2 - b^2 = 3 and 2ab = 4 placed b= 2/.a into first equation getting a^2-4/a^4=3 a^4 -3a^2 -4 =0This is a quadratic in a^2 which may be clean up through factoring giving a^2 = 4 or -a million. Reject the -a million as we'd like a real answer so x^2 = 4 and a = 2 or -2 b=2/a so b= a million or -a million therefore the sqrt of three+=4i is two+i or 2-i'm hoping this helped.

Answer 3

First, do a bit of algebra to isolate z. You get:
z = (1+5i)/(1+2i)

At this point, the trick is to multiply numerator and denominator by (1-2i). That turns the denominator into (1+2i)(1-2i), which turns into 1-4i^2 - and since i^2 = -1, that equals 1+4 = 5.

That should get you to where you can do the problem :-)

Answer 4

(i+z) - 3i(2-z) = iz +1
i+z - 6i +3iz = iz +1
...
z(1+2i) = 1+5i
z = (1+5i) / (1+2i)

Now comes the tricky part. Multiply top and bottom by (1-2i)

z = (1+5i)*(1-2i) / ((1+2i)*(1-2i))
= (1+5i) (1-2i) / 5
= (1 + 5i -2i +10) / 5
= (11+3i) / 5

Answer 5

group together all the terms without z
and all the terms with z

i+z-6i+3iz= iz+1
z(1+3i-i)=1-i+6i
z(1+2i)=1+5i
z=(1+5i)/(1+2i)
to get rid of the quotient youcan multiply and divide by 1-2i:
z=(1+5i)(1-2i)/(1+2i)(1-2i)
z=(11 +3i) /(1+4)
= 11/5 +3/5 i

Answer 6

not to be mean here but you have the beginning and you have the end just do some multiplication and arrangement ....

i'll do a lil for you :
(i+z) - 6i + 3iz = iz + 1
i - 6i + z + 2iz = 1
z + 2iz = 5i + 1 ..... now do some thinking here and solve for z

Answer 7

j+z-3j(2-z)=jz+1
multiply out to remove the ()
j+z-6j+3jz=jz+1
move stuff without z to the RHS
z+3jz-jz=+1-j+6j
collect
z(1+2j) = 1+5j
divide both sides by (1+2j)
z = (1+5j)/(1+2j)
now multiple the RHS by (1-2j)/(1-2j) -this will not change the value of the RHS, but, it will get rid of the j on the bottom
z = (1+5j)(1-2j)/(5) // because (1-2j)(1+2j) = 1+4 = 5
now multiply
z = (11+3j)/5 // because (1-2j)(1+5j) = 1 +5j -2j -10j*j = 11+3j
so
z = 11/5 + 3j/5
you should be careful about NOT putting the j on the bottom! as you did in your details section.

Answer 8

Just solve the equation for z as you'd normally do. Treat i as any other constant.