Express the complex number of z/z+3 =3−i the form a+bi

Mathematics

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Answer

−215−35i-\frac{21}{5} - \frac{3}{5}i

Step-by-step solution

5 steps

To express the complex number zz in the form a+bia+bi, given the equation zz+3=3−i\frac{z}{z+3} = 3-i:

  1. Rearrange the equation to solve for zz.

    zz+3=3−i\frac{z}{z+3} = 3-i

    Multiply both sides by (z+3)(z+3):

    z=(3−i)(z+3)z = (3-i)(z+3) z=3z+9−iz−3iz = 3z + 9 - iz - 3i

  2. Group terms containing zz on one side and constant terms on the other.

    z−3z+iz=9−3iz - 3z + iz = 9 - 3i −2z+iz=9−3i-2z + iz = 9 - 3i

    Factor out zz:

    z(−2+i)=9−3iz(-2+i) = 9 - 3i

  3. Isolate zz by dividing by (−2+i)(-2+i).

    z=9−3i−2+iz = \frac{9 - 3i}{-2 + i}

  4. Multiply the numerator and denominator by the conjugate of the denominator to express zz in the form a+bia+bi.

    The conjugate of −2+i-2+i is −2−i-2-i.

    z=9−3i−2+i×−2−i−2−iz = \frac{9 - 3i}{-2 + i} \times \frac{-2 - i}{-2 - i}

    Calculate the denominator:

    (−2+i)(−2−i)=(−2)2−(i)2=4−(−1)=4+1=5(-2+i)(-2-i) = (-2)^2 - (i)^2 = 4 - (-1) = 4+1 = 5

    Calculate the numerator:

    (9−3i)(−2−i)=9(−2)+9(−i)−3i(−2)−3i(−i)(9 - 3i)(-2 - i) = 9(-2) + 9(-i) - 3i(-2) - 3i(-i) =−18−9i+6i+3i2= -18 - 9i + 6i + 3i^2 =−18−3i+3(−1)= -18 - 3i + 3(-1) =−18−3i−3= -18 - 3i - 3 =−21−3i= -21 - 3i

    Now combine the numerator and denominator:

    z=−21−3i5z = \frac{-21 - 3i}{5}

  5. Write zz in the form a+bia+bi.

    z=−215−35iz = -\frac{21}{5} - \frac{3}{5}i

    The complex number is −215−35i-\frac{21}{5} - \frac{3}{5}i.

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