A-Level Mathematics May/June 2025 Q6: Find the complex numbers z for which (z+4)/(z+4i) is real and |z|= √10. Give your answe…

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  Let $z = x+iy$. The condition that $\frac{z+4}{z+4i}$ is real is considered first.

  Substitute $z=x+iy$ into the expression: $\frac{(x+iy)+4}{(x+iy)+4i} = \frac{(x+4)+iy}{x+i(y+4)}$

  To make the denominator real, multiply the numerator and denominator by the conjugate of the denominator, which is x-i(y+4): $\frac{((x+4)+iy)(x-i(y+4))}{(x+i(y+4))(x-i(y+4))} = \frac{(x+4)x - i(x+4)(y+4) + ixy + y(y+4)}{x^2 + (y+4)^2}$

  For the expression to be real, the imaginary part of the numerator must be zero. $\text{Im}(\text{numerator}) = xy - (x+4)(y+4) = 0$

  Expanding and simplifying this equation: $xy - (xy + 4x + 4y + 16) = 0$ $-4x - 4y - 16 = 0$ $x + y + 4 = 0 \implies x+y = -4 \quad (*)$

  The second condition is $|z| = \sqrt{10}$. Since $z=x+iy$, we have $|z| = \sqrt{x^2+y^2}$. $\sqrt{x^2+y^2} = \sqrt{10} \implies x^2+y^2 = 10 \quad (**)$

  Now we solve the simultaneous equations () and (**). From (), we have $y = -4-x$.

  Substitute this into (**): $x^2 + (-4-x)^2 = 10$ $x^2 + (16 + 8x + x^2) = 10$ $2x^2 + 8x + 6 = 0$ $x^2 + 4x + 3 = 0$

  Factorising the quadratic equation: $(x+1)(x+3) = 0$ This gives two possible values for x: $x=-1$ or $x=-3$.

  Find the corresponding values for y using $y=-4-x$:

  Case 1: If $x=-1$, then $y = -4 - (-1) = -3$. This gives the complex number $z = -1 - 3i$.

  Case 2: If $x=-3$, then $y = -4 - (-3) = -1$. This gives the complex number $z = -3 - i$.

  The two complex numbers are $z = -3-i$ and $z = -1-3i$.

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  How the marks are awarded

  • M1 — Correctly setting up the fraction with z=x+iy and multiplying both the numerator and denominator by the conjugate of the denominator, which is x - i(y+4).
  • DM1 — Expanding the numerator and equating its imaginary part to zero. The imaginary part is correctly identified as xy - (x+4)(y+4). This mark is dependent on the first M1.
  • A1 — Correctly simplifying the equation from the imaginary part to obtain the linear relationship x + y = -4 or an equivalent form.
  • DM1 — Using the modulus condition |z|=√10 to get x²+y²=10 and substituting the linear relationship (e.g., y=-x-4) to form a correct quadratic equation in one variable, such as x²+$4x+3=0$.
  • A1 — Solving the quadratic and finding one of the correct solutions in the required form, e.g., z = -3-i.
  • A1 — Finding the second correct solution, z = -1-3i, and providing only these two solutions.

  Common mistakes

  • Using an incorrect conjugate for the denominator, such as x+iy-4i or z*-4i, instead of the conjugate of the entire denominator, x-i(y+4).
  • Making algebraic errors when expanding the numerator, particularly with the signs in -(x+4)(y+4), leading to an incorrect linear equation.
  • Correctly finding the pairs of values (x=-1, y=-3) and (x=-3, y=-1) but failing to write the final answers as complex numbers in the required form z = x+iy.
  • Mistakes in solving the simultaneous equations, for example, incorrectly expanding (-4-x)² or errors in factorising the final quadratic.

  Examiner tip: To find conditions for a complex fraction to be purely real or imaginary, always start by multiplying the numerator and denominator by the conjugate of the denominator.

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