Complex Numbers and Quadratic Equations: questions with answers

Im z is the REAL number b, not the term ib. Both parts of a complex number are real numbers; what makes the number complex is the way they are combined. Note that when b is negative the minus sign belongs to the answer — Im z is 6 and it does not become positive.

Equality of $a+bi$ and $c+di$ gives $a=c$ and $b=d$: two real component equalities. This does not mean that they are always independent equations; a component equality can already be an identity. The comparison still has a real part and an imaginary part.

The gap between the two is exactly -bd, the contribution of the i² term, and it vanishes only when one of the imaginary parts is 0. Here both are non-zero in every case, so the two quantities never agree -- which is the point the exercise is making.

This is what 'enlarging the number system' means: the real numbers did not go anywhere. Every one of them sits inside the complex numbers as a number whose imaginary part happens to be 0, and the chapter writes its own answers that way — 3 + i0, 0 + i0 — whenever it wants the standard form.

A number times its conjugate is the SQUARE of the modulus, so one more step is needed than the given supplies. And x and y are never needed separately: the product already contains everything the modulus depends on, which is exactly what makes this factorisation worth recognising.

Each part can independently have either sign. Squaring removes those signs, so all four combinations have the same modulus.

The minus sign in front of the bracket reaches BOTH parts of z2, not just the one written first. This is the ordinary bracket rule, but it is unforgiving here because half the number lives behind the i -- and when 5 is negative, subtracting it makes the imaginary part grow.

Multiplying numerator and denominator by the same nonzero number preserves the quotient. The displayed numerator-conjugate step is therefore an equivalent expression, but it leaves a complex denominator. Direct rationalisation needs the denominator’s conjugate 4 - i3, giving real denominator 25. The first line that fails this stated purpose is Line 2; it is not a false equality.

Reduce each inner exponent modulo four, add the resulting complex numbers, and then take the outer power.

The binomial expansion is valid. The first incorrect equality treats i² as 1, making the last squared term positive instead of negative.

The difference-of-squares pattern is right, but the square being subtracted carries an i: (ib)² is -b², so subtracting it ADDS b². This is why the denominator is always positive and the inverse always exists for a non-zero z -- whereas a² - b² would be 7 here, and at a = b it would be 0.

The horizontal coordinate becomes the real part and the vertical one becomes the imaginary part -- as an ordinary real number multiplying i, not as an i of its own. Treating the vertical coordinate as 4i would make the number 3 + i(4i), which is the real number -1.

The triangle inequality gives |z1 + z2| ≤ |z1| + |z2|. These pairs have different directions, so the gap is positive.

The principal square root of -4 is 2i. Subtract the candidate by subtracting its real and imaginary parts separately.

Conjugation reflects the point across the real axis. The first and fourth quadrants exchange, and the second and third quadrants exchange.