Let z=x+yi, then∣x+i(3+y)∣=∣(x+5)+i(y−2)∣By definition of magnitude we have thatx2+(3+y)2=(x+5)2+(y−2)2=10x−10y=−20⟹x−y=−2−(1)Also for the second equation, we have that ∣x+(y−4)i∣=∣x+(2+y)i∣=x2+(y−4)2=x2+(2+y)2Simplifying the equation, we have thaty=1From eqn(1), we have that x=-2+y⟹x=−1Therefore z=−1+i
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