Q.If $(x + 1,\ y - 2) = (3, 1)$, find the values of $x$ and $y$.

Concept and Intuition

When an equation mixes $z$ and $\bar z$, polar form $z=re^{i\theta}$ ($r\ge 0$) is the natural tool because $\bar z = re^{-i\theta}$ and powers of $z$ are easy in this form. We must not forget the trivial solution $z=0$, which polar substitution can silently drop if $r$ is divided out.

Step-by-Step Solution

1. Let $z=re^{i\theta}$, $r\ge 0$. Then $\bar z = re^{-i\theta}$, and the equation becomes

2. Case $r=0$: gives $z=0$, which indeed satisfies $0+0=0$. This is one solution.
3. Case $r>0$: divide the equation by $re^{-i\theta}$ (valid since $r\neq0$):

4. Taking modulus: $r^2=1\Rightarrow r=1$ (since $r>0$).
5. Taking argument: $e^{i4\theta}=-1=e^{i(\pi+2k\pi)}$, so $4\theta=\pi+2k\pi$, i.e. $\theta = \dfrac{\pi}{4}+\dfrac{k\pi}{2}$ for $k=0,1,2,3$ giving four distinct values of $\theta$ in $[0,2\pi)$: $\pi/4,\ 3\pi/4,\ 5\pi/4,\ 7\pi/4$.
6. This gives 4 distinct nonzero solutions on the unit circle.
7. Total distinct solutions $= 4\text{ (nonzero)} + 1\text{ (zero)} = 5$.

Common Mistakes

• Dividing by $\bar z$ or $z$ too early and losing the $z=0$ solution.
• Forgetting that $e^{i4\theta}=-1$ has 4 solutions for $\theta$ in a full period, not just 1.

ANSWER: D

Concept and Intuition

If a polynomial has real coefficients and a non-real complex number is a root, its complex conjugate must also be a root (the Conjugate Root Theorem). This instantly gives the second root without any further work, and then Vieta's formulas (sum/product of roots) hand us $a$ and $b$ directly.

Step-by-Step Solution

1. $x^2+ax+b=0$ has real coefficients $a,b$, and $(3+i)$ is a root.
2. By the conjugate root theorem, $(3-i)$ is also a root.
3. Sum of roots $=(3+i)+(3-i)=6$.
4. For $x^2+ax+b=0$, sum of roots $=-a$. So $-a=6 \Rightarrow a=-6$.
5. (Product of roots $=(3+i)(3-i)=9+1=10=b$, confirming consistency, though only $a$ is asked.)

Common Mistakes

• Sign error: forgetting sum of roots is $-a$, not $a$, for $x^2+ax+b=0$.
• Forgetting that real coefficients force the conjugate to also be a root, and instead trying to solve for $a,b$ using only one root (underdetermined).

ANSWER: D

Concept and Intuition

When an equation between complex expressions involves only real unknowns a,b, we equate real parts and imaginary parts separately.

Step-by-Step Solution

1. Given: $4a + i(3a-b) = b - 6i$, a,b real.
2. Real parts: $4a = b$.
3. Imaginary parts: $3a-b = -6$.
4. Substitute $b=4a$: $3a-4a=-6 \Rightarrow a=6$.
5. $b = 4a = 24$.
6. $z = a+\dfrac{b}{4}i = 6+6i$.
7. $|z| = \sqrt{36+36}=6\sqrt2$.
8. $\dfrac{|z|}{a} = \sqrt2$.

Common Mistakes

• Forgetting b in z=a+(b/4)i is the same b found from the original equation.
• Sign slip equating imaginary parts (RHS has -6i).

ANSWER: C

Concept and Intuition

If $z_1=z_2^*$ (complex conjugates), then $\mathrm{Re}(z_1)=\mathrm{Re}(z_2)$ and $\mathrm{Im}(z_1)=-\mathrm{Im}(z_2)$. Here $z_1=-3+ix^2y$ has real part $-3$ and imaginary part $x^2y$; $z_2=x^2+y+4i$ has real part $x^2+y$ and imaginary part $4$.

Step-by-Step Solution

1. Real parts equal: $-3 = x^2+y \Rightarrow y=-3-x^2$.
2. Imaginary parts opposite: $x^2y = -4$.
3. Substitute: $x^2(-3-x^2)=-4 \Rightarrow -3x^2-x^4=-4 \Rightarrow x^4+3x^2-4=0$.
4. Let $u=x^2$: $u^2+3u-4=0 \Rightarrow (u+4)(u-1)=0 \Rightarrow u=1$ or $u=-4$.
5. Since $u=x^2\ge0$, reject $u=-4$; take $u=1 \Rightarrow x^2=1 \Rightarrow x=\pm1$.

Common Mistakes

• Forgetting to discard the negative value of $u=x^2=-4$ (not possible for real $x$).
• Mixing up which expression's imaginary part must be negated.

ANSWER: B

Concept and Intuition

Since $x,y$ are constrained to be real, an equation like $5z_1+xz_2+yz_3=0$ (a complex equation) splits into two independent real equations — one from the real parts and one from the imaginary parts.

Step-by-Step Solution

1. $z_3 = \dfrac{z_1+z_2}{2} = \dfrac{(2+3i)+(4-5i)}{2} = \dfrac{6-2i}{2}=3-i$.
2. Substitute: $5(2+3i)+x(4-5i)+y(3-i)=0$.
3. Expand: $(10+15i)+(4x-5xi)+(3y-yi)=0$.
4. Real part: $10+4x+3y=0$. Imaginary part: $15-5x-y=0 \Rightarrow y=15-5x$.
5. Substitute into real equation: $10+4x+3(15-5x)=0 \Rightarrow 10+4x+45-15x=0 \Rightarrow 55-11x=0 \Rightarrow x=5$.
6. Then $y=15-25=-10$.
7. $x+y = 5+(-10) = -5$.

Common Mistakes

• Forgetting that $x,y$ being real means the complex equation must be split into two real equations rather than solved as one complex unknown.

ANSWER: A

We start with the equation

Our goal is to find real numbers $x$ and $y$ that satisfy this. Since the right-hand side is purely imaginary, the sum of the two fractions must also be purely imaginary. The natural approach is to rewrite each fraction in standard $a+bi$ form, then equate real and imaginary parts.

Why this works:

Complex numbers are equal iff their real parts are equal and their imaginary parts are equal. So if we simplify the left side to $A + Bi$, we must have $A = 0$ (since the right side has no real part) and $B = 1$. This gives two equations in $x$ and $y$.

Let’s proceed step by step.

1. Rationalise the first fraction Multiply numerator and denominator by the conjugate of $3+i$, which is $3-i$:

2. Rationalise the second fraction Multiply numerator and denominator by the conjugate of $3-i$, which is $3+i$:

3. Add the two fractions Since both have denominator 10, we add the numerators:

4. Expand the numerator First term: $(x-1)(3-i) = 3(x-1) - i(x-1)$. Second term: $(y-1)(3+i) = 3(y-1) + i(y-1)$. Sum:

Simplify the real part: $3(x-1+y-1) = 3(x+y-2)$.

Simplify the imaginary part: $-(x-1) + (y-1) = -x+1 + y-1 = y - x$.

So the left side becomes:

5. Multiply both sides by 10

6. Equate real and imaginary parts The left side is $3(x+y-2) + i(y-x)$. The right side is $0 + 10i$. Therefore:

7. Solve the system From $x+y=2$ and $y-x=10$, add the equations: $(x+y)+(y-x)=2+10 \Rightarrow 2y=12 \Rightarrow y=6$. Then $x = 2 - y = 2 - 6 = -4$.

So we have $x = -4$ and $y = 6$. That means $x < 0$ and $y > 0$.

Thus the correct choice is (B).

Concept and Intuition

When a complex number is expressed as a power of a simpler complex number, expanding the binomial and separating real/imaginary parts converts the problem into ordinary algebra in $a$ and $b$ — this is the standard trick behind De Moivre's theorem applications like this one.

Step-by-Step Solution

1. $(a-ib)^3 = a^3 - 3a^2(ib) + 3a(ib)^2 - (ib)^3$.
2. $(ib)^2=-b^2$, $(ib)^3=-ib^3$, so this becomes $a^3 - 3ia^2b - 3ab^2 + ib^3 = (a^3-3ab^2) + i(b^3-3a^2b)$.
3. Given $(x-iy) = (a-ib)^3$, match real and imaginary parts: $x = a^3-3ab^2$ and $-y = b^3-3a^2b$, i.e. $y = 3a^2b-b^3$.
4. Compute $\dfrac{x}{2a} = \dfrac{a^3-3ab^2}{2a} = \dfrac{a^2-3b^2}{2}$ and $\dfrac{y}{2b} = \dfrac{3a^2b-b^3}{2b} = \dfrac{3a^2-b^2}{2}$.
5. Sum: $\dfrac{a^2-3b^2}{2}+\dfrac{3a^2-b^2}{2} = \dfrac{4a^2-4b^2}{2} = 2(a^2-b^2)$.

Common Mistakes

• Sign error when separating the imaginary part (forgetting the overall minus sign linking $-y$ to the expansion).
• Arithmetic slip combining the two fractions before simplifying.

ANSWER: A

Concept and Intuition

Two complex numbers $z_1, z_2$ are conjugates of each other precisely when $z_2 = \overline{z_1}$. Writing both numbers in real+imaginary form and equating real parts to real parts, imaginary parts to imaginary parts, turns a complex condition into two ordinary trigonometric equations that must hold together.

Step-by-Step Solution

1. Let $z_1=\sin x+i\cos 2x$ and $z_2=\cos x-i\sin 2x$. "Conjugate to each other" means $z_2=\overline{z_1}$.
2. $\overline{z_1} = \sin x - i\cos 2x$.
3. Setting $z_2=\overline{z_1}$: $\cos x - i\sin 2x = \sin x - i\cos 2x$.
4. Equate real parts: $\cos x = \sin x \;\Rightarrow\; \tan x = 1 \;\Rightarrow\; x = n\pi+\dfrac{\pi}{4}$.
5. Equate imaginary parts: $-\sin 2x = -\cos 2x \;\Rightarrow\; \sin 2x=\cos 2x$.
6. Now check whether $x=n\pi+\frac{\pi}{4}$ can also satisfy step 5: $2x = 2n\pi+\frac{\pi}{2}$, so $\sin 2x = \sin\left(\frac{\pi}{2}\right)=1$ and $\cos 2x=\cos\left(\frac{\pi}{2}\right)=0$ for every integer $n$. Since $1\neq 0$, condition 5 is never satisfied.
7. Since the value of $x$ forced by the real-part condition never satisfies the imaginary-part condition, there is no $x$ that makes the two numbers conjugate.

Common Mistakes

• Solving the two trig equations $\tan x=1$ and $\tan 2x=1$ independently and reporting their individual solution sets, instead of checking that a single $x$ must satisfy both simultaneously.
• Forgetting that "conjugate to each other" is symmetric, so it's enough to check $z_2=\overline{z_1}$ once (no need to separately check $z_1=\overline{z_2}$, which is the same condition).

ANSWER: B

Note on the printed stem: the second numerator must read $(2-3i)y + i$. As printed (with the $+\,i$ dropped in transcription) the equation gives the non-listed value $x+y = \tfrac{49}{23}$; the solution below works the intended equation.

Concept and Intuition

Two complex numbers are equal iff their real parts are equal and their imaginary parts are equal. That single fact turns one complex equation into two simultaneous real equations — which is exactly what we need to pin down the two real unknowns $x$ and $y$.

The only technical step is clearing the complex denominators, done by multiplying numerator and denominator by the conjugate:

Step-by-Step Solution

1. Clear the first denominator. Numerator $(1+i)x - 2i = x + i(x-2)$. Multiply by $(3-i)$:

2. Clear the second denominator. Numerator $(2-3i)y + i = 2y + i(1-3y)$. Multiply by $(3+i)$:

3. Assemble (both fractions carry the same denominator $10$, and the RHS is $i$, i.e. $10i$ after multiplying through by $10$):

4. Equate parts.

5. Solve. Doubling the second: $4x - 14y = 26$. Subtracting from the first:

6. $x + y = 3 + (-1) = \boxed{2}$.

Common Mistakes

• Forgetting $i^2 = -1$ when expanding, which flips the sign of the last term in each product.
• Equating the two fractions separately to $i/2$ each — the equation only constrains the sum.
• Rationalising with the wrong conjugate: $3+i$ pairs with $3-i$ (and vice versa), never with itself.

ANSWER: D

Concept and Intuition

For any real $x$, the numbers $3-4ix$ and $3+4ix$ are complex conjugates of each other. The ratio

of a nonzero complex number to its own conjugate, $\bar w/w$, always has modulus $1$ (since

$|\bar w|=|w|$). So no matter what real value $x$ takes, the given ratio is automatically a

modulus-1 complex number.

Step-by-Step Solution

1. For real $x$, let $w = 3+4ix$. Then $\bar w = 3-4ix$.
2. The given ratio is $\dfrac{\bar w}{w} = \alpha - i\beta$.
3. Take modulus of both sides: $\left|\dfrac{\bar w}{w}\right| = \dfrac{|\bar w|}{|w|} = \dfrac{|w|}{|w|} = 1$ (since $|\bar w|=|w|$ always).
4. So $|\alpha - i\beta| = 1 \Rightarrow \sqrt{\alpha^2+\beta^2} = 1 \Rightarrow \alpha^2+\beta^2=1$.
5. This holds for any real $x$ (as long as $w\ne0$), which is exactly why the question says "a real value of $x$ will satisfy this equation if..." — the condition on $\alpha,\beta$ is simply that they come from a modulus-1 ratio.

Common Mistakes

• Trying to expand the ratio algebraically into real/imaginary parts and separately squaring and subtracting — this also works but is far more error-prone than recognising the conjugate-ratio modulus-1 shortcut.

ANSWER: C

Concept and Intuition

Rather than expanding $(a+bi)^3$ in full real/imaginary parts (messy), it's far cleaner to write $z=a+bi$ so the condition becomes the compact complex equation $z^3=\bar z$. Using the fact that for any complex number $|z\bar z|=|z|^2$ and $\bar z = 1/z$ when $|z|=1$, we can reduce this to a root-of-unity problem.

Step-by-Step Solution

1. Let $z=a+bi$. The condition $(a+bi)^3=a-bi$ becomes $z^3=\bar z$.
2. Case $z=0$: trivially satisfies $z^3=\bar z$ (both sides are 0). This gives $(a,b)=(0,0)$ — one solution.
3. Case $z\neq 0$: Take the modulus of both sides: $|z^3|=|\bar z| \Rightarrow |z|^3=|z|$. Since $|z|\neq 0$, divide by $|z|$: $|z|^2=1 \Rightarrow |z|=1$.
4. Since $|z|=1$, we have $\bar z = \dfrac{1}{z}$. Substituting into $z^3=\bar z$: $z^3=\dfrac1z \Rightarrow z^4=1$.
5. $z^4=1$ has exactly 4 roots on the unit circle: $z=1,\,-1,\,i,\,-i$ — all automatically satisfy $|z|=1$, consistent with our case assumption.
6. Verify each quickly: $z=1$: $1^3=1=\bar1$ ✓. $z=-1$: $(-1)^3=-1=\overline{-1}$ ✓. $z=i$: $i^3=-i=\bar i$ ✓. $z=-i$: $(-i)^3=i=\overline{-i}$ ✓. All 4 check out.
7. Total ordered pairs $(a,b)$: the zero solution plus the 4 roots of unity = $1+4=5$.

Common Mistakes

• Forgetting the $z=0$ solution and stopping at 4.
• Expanding $(a+bi)^3$ term-by-term and making a sign/algebra slip in separating real and imaginary parts — the modulus trick avoids this entirely.
• Assuming $\bar z=1/z$ holds in general (it only holds when $|z|=1$, which is precisely the case we've derived).

ANSWER: D