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Solve for z: z / (z+2) = 2−i

Mathematics

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Answer

-3 - i

Step-by-step solution

10 steps

  1. Isolate zz by multiplying both sides by (z+2)(z+2).

    We are given the equation:

    zz+2=2−i\frac{z}{z+2} = 2-i

    Multiply both sides by (z+2)(z+2):

    z=(2−i)(z+2)z = (2-i)(z+2)

  2. Expand the right side of the equation.

    z=2(z)+2(2)−i(z)−i(2)z = 2(z) + 2(2) - i(z) - i(2) z=2z+4−iz−2iz = 2z + 4 - iz - 2i

  3. Gather all terms containing zz on one side and constant terms on the other side.

    Subtract 2z2z and add iziz to both sides:

    z−2z+iz=4−2iz - 2z + iz = 4 - 2i −z+iz=4−2i-z + iz = 4 - 2i

  4. Factor out zz from the terms on the left side.

    z(−1+i)=4−2iz(-1 + i) = 4 - 2i

  5. Solve for zz by dividing both sides by (−1+i)(-1+i).

    z=4−2i−1+iz = \frac{4 - 2i}{-1 + i}

  6. Rationalize the denominator by multiplying the numerator and denominator by the conjugate of the denominator.

    The conjugate of −1+i-1+i is −1−i-1-i.

    z=4−2i−1+i×−1−i−1−iz = \frac{4 - 2i}{-1 + i} \times \frac{-1 - i}{-1 - i}

  7. Perform the multiplication in the numerator.

    (4−2i)(−1−i)=4(−1)+4(−i)−2i(−1)−2i(−i)(4 - 2i)(-1 - i) = 4(-1) + 4(-i) - 2i(-1) - 2i(-i) =−4−4i+2i+2i2= -4 - 4i + 2i + 2i^2

    Since i2=−1i^2 = -1:

    =−4−2i+2(−1)= -4 - 2i + 2(-1) =−4−2i−2= -4 - 2i - 2 =−6−2i= -6 - 2i

  8. Perform the multiplication in the denominator.

    (−1+i)(−1−i)=(−1)2−(i)2(-1 + i)(-1 - i) = (-1)^2 - (i)^2 =1−(−1)= 1 - (-1) =1+1= 1 + 1 =2= 2

  9. Combine the simplified numerator and denominator.

    z=−6−2i2z = \frac{-6 - 2i}{2}

  10. Simplify the expression to the form a+bia+bi.

    z=−62−2i2z = \frac{-6}{2} - \frac{2i}{2} z=−3−iz = -3 - i

    The value of zz in the form a+bia+bi is −3−i\boxed{-3 - i}.

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