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Algebra Difficulty 2.9 Junior Find the answer

Find the complex number zz that satisfies the equation: z(1+i)+i=0z(1+i)+i=0.

Pick one

Solution

Given z(1+i)+i=0z(1+i)+i=0, we first isolate zz:

z=i1+iz = \frac{-i}{1+i}

To simplify this expression, we can multiply the numerator and the denominator by the conjugate of the denominator:

z=i(1i)(1+i)(1i)z = \frac{-i(1-i)}{(1+i)(1-i)}

Recall that the product of a complex number and its conjugate is the square of its magnitude, hence (1+i)(1i)=12+12=2(1+i)(1-i)=1^2+1^2=2:

z=1i2z = \frac{-1-i}{2}

Finally, separate real and imaginary parts to obtain:

z=1212iz = -\frac{1}{2} - \frac{1}{2}i

Therefore, the correct answer is:
\boxed{B: -\frac{1}{2} - \frac{1}{2}i}

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.