Porisma

Worked solution

Use both conjugate constraints to determine z

Find the possible values of zz in the form x+iyx+iy, where x,y∈Rx,y\in\mathbb R, satisfying both zz∗=35+45i\dfrac{z}{z^*}=\dfrac35+\dfrac45i and zz∗=5zz^*=5. marks

[2013/NYJC/I/6(a)]

CJC · Review problems · Question 1

Answer

z=2+iz=2+i or z=−2−iz=-2-i.

Reduce to a square

Since zz∗=5zz^*=5, z≠0z\ne0 and z∗=5/zz^*=5/z. Hence z/z∗=z2/5=3/5+(4/5)iz/z^*=z^2/5=3/5+(4/5)i, giving z2=3+4iz^2=3+4i.

How did this part go?

Compare real and imaginary parts

Let z=x+iyz=x+iy. Then x2−y2=3x^2-y^2=3 and 2xy=42xy=4, so x≠0x\ne0 and y=2/xy=2/x. Substitute: x2−4/x2=3x^2-4/x^2=3, hence x4−3x2−4=0x^4-3x^2-4=0, or (x2−4)(x2+1)=0(x^2-4)(x^2+1)=0. Reject x2=−1x^2=-1 because xx is real. The paired possibilities are (x,y)=(2,1)(x,y)=(2,1) or (−2,−1)(-2,-1).

How did this part go?

Return to both original conditions

Therefore z=2+iz=2+i or z=−2−iz=-2-i. For either sign, zz∗=4+1=5zz^*=4+1=5 and z/z∗=z2/(zz∗)=(3+4i)/5z/z^*=z^2/(zz^*)=(3+4i)/5. Thus neither original constraint has been lost.

How did this part go?

Common mistakes

The signs of x and y are linked by xy=2. The two roots are a pair of opposites, not four independent sign choices.

Solution notes

Source filename says 2026, but its printed header says 2025/JC2; both are retained without silently changing the source identity.

Read the hint