Blackmancan
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Solve for z:
(2z/2+i) + 3-2i=(1-i)z
Thanks in advance
(2z/2+i) + 3-2i=(1-i)z
Thanks in advance
Complex Help (1 Viewer)
- Thread starter Blackmancan
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sorry I don't have a pen and paper, but let z=x+iy then you substitute and simplify, then you equate the real and imaginary. Btw u can not have an imaginary as the denominator so you must rationalise.
A simpler way is to factor out z and then solve for z.
Yeh true but I recommend using the substitution method because it is useful for almost any other situation whereas the factorisation may not work sometimes, but yes you can
bleakarcher
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z=[11+5i]/2
we want to make z the subject as follows;
= (1-i)z)
so first we dont want to have any fractions so we times everything by (2+i)
(3-2i)= (2+i)(1-i)z)
now we take the 2z to the other side so we can factor it out:
= (2+i-2i-i^{2})z -2z)
simplyfying down and taking z out this becomes:
= ((3-i) -2)z)
making z the subject now:
relising the denom this becomes:
(1+i)}{(1-i)(1+i)})
which simplifies to:
so first we dont want to have any fractions so we times everything by (2+i)
now we take the 2z to the other side so we can factor it out:
simplyfying down and taking z out this becomes:
making z the subject now:
relising the denom this becomes:
which simplifies to: