Questions and Answers Section

Example 1: Simplifying Powers of Iota

Question: Find the value of $\frac{i^{4n+1} - i^{4n-1}}{2}$.

Solution: We can factor out $i^{4n}$ from the numerator. Using the property $i^{4n} = (i^4)^n = 1^n = 1$.

$$\frac{i^{4n} \cdot i^1 - i^{4n} \cdot i^{-1}}{2} = \frac{1 \cdot i - 1 \cdot \frac{1}{i}}{2}$$

Since $1/i = -i$, the expression becomes:

$$\frac{i - (-i)}{2} = \frac{2i}{2} = i$$

Example 2: Equality of Complex Numbers

Question: Find the real number $x$ if $(x-2i)(1+i)$ is purely imaginary.

Solution: First, expand the expression: $(x-2i)(1+i) = x + xi - 2i - 2i^2 = x + (x-2)i - 2(-1) = (x+2) + (x-2)i$. For a complex number to be purely imaginary, its real part must be zero. $x+2 = 0 \implies x = -2$.

Example 3: Finding Modulus and Argument

Question: Find the modulus and principal argument of $z = \frac{1+i}{1-i}$.

Solution: First, simplify z:

$$z = \frac{1+i}{1-i} = \frac{(1+i)(1+i)}{(1-i)(1+i)} = \frac{1+2i+i^2}{1^2+1^2} = \frac{1+2i-1}{2} = \frac{2i}{2} = i$$

So, $z = 0+1i$.

• Modulus: $|z| = \sqrt{0^2+1^2} = 1$.
• Argument: The point (0,1) lies on the positive imaginary axis. The angle it makes with the positive real axis is $\pi/2$. So, $\arg(z) = \pi/2$.

Example 4: Using Properties of Conjugates

Question: If $(x+iy)^{1/3} = a+ib$, show that $\frac{x}{a} + \frac{y}{b} = 4(a^2-b^2)$.

Solution: We are given $x+iy = (a+ib)^3$. $x+iy = a^3 + 3a^2(ib) + 3a(ib)^2 + (ib)^3 = (a^3 - 3ab^2) + i(3a^2b - b^3)$. Equating real and imaginary parts: $x = a^3 - 3ab^2 = a(a^2-3b^2)$ $y = 3a^2b - b^3 = b(3a^2-b^2)$ From these, $\frac{x}{a} = a^2-3b^2$ and $\frac{y}{b} = 3a^2-b^2$. Adding them gives:

$$\frac{x}{a} + \frac{y}{b} = (a^2-3b^2) + (3a^2-b^2) = 4a^2-4b^2 = 4(a^2-b^2)$$

Example 5: Solving an Equation with Modulus

Question: Find the complex number z if $|z|+z = 2+i$.

Solution: Let $z=x+iy$. The equation becomes $\sqrt{x^2+y^2} + x+iy = 2+i$. Equating the imaginary parts, we get $y=1$. Equating the real parts: $\sqrt{x^2+y^2} + x = 2$. Substitute $y=1$: $\sqrt{x^2+1} + x = 2 \implies \sqrt{x^2+1} = 2-x$. Squaring both sides: $x^2+1 = (2-x)^2 = 4-4x+x^2$. $1 = 4-4x \implies 4x=3 \implies x=3/4$. So, the complex number is $z = \frac{3}{4}+i$.

Example 6: Using De Moivre's Theorem

Question: Evaluate $(\cos(\pi/15) + i\sin(\pi/15))^5$.

Solution: By De Moivre's Theorem, $(\cos\theta + i\sin\theta)^n = \cos(n\theta) + i\sin(n\theta)$.

$$(\cos(\pi/15) + i\sin(\pi/15))^5 = \cos(5 \cdot \pi/15) + i\sin(5 \cdot \pi/15)$$

$$= \cos(\pi/3) + i\sin(\pi/3) = \frac{1}{2} + i\frac{\sqrt{3}}{2}$$

Example 7: Finding Cube Roots of Unity

Question: Solve the equation $z^3-1=0$.

Solution: $z^3-1 = (z-1)(z^2+z+1)=0$. One root is $z=1$. The other two roots come from $z^2+z+1=0$. Using the quadratic formula: $z = \frac{-1 \pm \sqrt{1^2-4(1)(1)}}{2} = \frac{-1 \pm \sqrt{-3}}{2} = -\frac{1}{2} \pm i\frac{\sqrt{3}}{2}$. The three cube roots of unity are $1, \omega, \omega^2$, where $\omega = -\frac{1}{2} + i\frac{\sqrt{3}}{2}$.

Example 8: Properties of Cube Roots of Unity

Question: If $\omega$ is a complex cube root of unity, find the value of $(1+\omega)(1+\omega^2)(1+\omega^4)(1+\omega^5)$.

Solution: Use properties $1+\omega+\omega^2=0$ and $\omega^3=1$. $1+\omega = -\omega^2$ and $1+\omega^2 = -\omega$. Also, $\omega^4=\omega$ and $\omega^5=\omega^2$. The expression becomes:

$$(-\omega^2)(-\omega)(1+\omega)(1+\omega^2) = (\omega^3)(-\omega^2)(-\omega) = (1)(\omega^3) = 1$$

Example 9: Locus of a Point

Question: Find the Cartesian equation of the locus of z satisfying $|z-4| = 3$.

Solution: Let $z=x+iy$. The equation is $|(x+iy)-4|=3 \implies |(x-4)+iy|=3$. By definition of modulus, $\sqrt{(x-4)^2 + y^2} = 3$. Squaring both sides gives $(x-4)^2+y^2=9$. This is the equation of a circle with center (4,0) and radius 3.

Example 10: Solving a Quadratic Equation

Question: Solve $x^2 - 4x + 13 = 0$.

Solution: Using the quadratic formula, $x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$. $D = (-4)^2 - 4(1)(13) = 16 - 52 = -36$. $x = \frac{4 \pm \sqrt{-36}}{2} = \frac{4 \pm 6i}{2} = 2 \pm 3i$.

Example 11: Using Triangle Inequality

Question: If $|z|=3$, find the range of values for $|z+4|$.

Solution: By the triangle inequality, $||z|-|4|| \le |z+4| \le |z|+|4|$. Substituting $|z|=3$ and $|4|=4$: $|3-4| \le |z+4| \le 3+4$. $1 \le |z+4| \le 7$. The values lie in the interval [1, 7].

Example 12: Argument Properties

Question: If $z_1=1+i$ and $z_2=\sqrt{3}+i$, find $\arg(z_1z_2)$.

Solution: We use the property $\arg(z_1z_2) = \arg(z_1) + \arg(z_2)$.

• For $z_1=1+i$, $\arg(z_1) = \tan^{-1}(1/1) = \pi/4$.
• For $z_2=\sqrt{3}+i$, $\arg(z_2) = \tan^{-1}(1/\sqrt{3}) = \pi/6$.

$$\arg(z_1z_2) = \frac{\pi}{4} + \frac{\pi}{6} = \frac{3\pi+2\pi}{12} = \frac{5\pi}{12}$$

Example 13: Finding Square Roots

Question: Find the square roots of $-5 + 12i$.

Solution: Let $\sqrt{-5+12i}=x+iy$. Then $-5+12i = (x^2-y^2)+2ixy$. $x^2-y^2=-5$ and $2xy=12 \implies xy=6$. $(x^2+y^2)^2 = (x^2-y^2)^2+(2xy)^2 = (-5)^2+12^2 = 25+144=169$. So, $x^2+y^2=13$. Adding $x^2+y^2=13$ and $x^2-y^2=-5$ gives $2x^2=8 \implies x=\pm 2$. Subtracting gives $2y^2=18 \implies y=\pm 3$. Since $xy=6$ is positive, x and y have the same sign. The roots are $2+3i$ and $-2-3i$.

Example 14: Geometric Interpretation

Question: If $z_1, z_2, z_3$ are the vertices of an equilateral triangle, prove that $z_1^2+z_2^2+z_3^2 = z_1z_2+z_2z_3+z_3z_1$.

Solution: For an equilateral triangle, the rotation of one side to another is by an angle of $\pi/3$. A key property derived from this is that if $z_1, z_2, z_3$ are vertices of an equilateral triangle, then $(z_1-z_2)^2 + (z_2-z_3)^2 + (z_3-z_1)^2 = 0$. Expanding this gives the desired result after simplification. Another approach: A known condition for vertices of an equilateral triangle is $\frac{z_1-z_2}{z_3-z_2} = \frac{z_3-z_1}{z_2-z_1}$. Cross-multiplying and simplifying leads to the result.

Example 15: Forming Equation with Complex Roots

Question: Find the quadratic equation with real coefficients if one root is $\frac{1}{2-i}$.

Solution: First, simplify the root: $z_1 = \frac{1}{2-i} = \frac{2+i}{2^2+1^2} = \frac{2}{5} + i\frac{1}{5}$. Since coefficients are real, the other root is the conjugate: $z_2 = \frac{2}{5} - i\frac{1}{5}$. Sum = $z_1+z_2 = 4/5$. Product = $z_1z_2 = |z_1|^2 = (2/5)^2+(1/5)^2 = 5/25 = 1/5$. The equation is $x^2 - (4/5)x + 1/5 = 0$, or $5x^2 - 4x + 1 = 0$.

Example 16: Complex Numbers and GP

Question: If the cube roots of unity are $1, \omega, \omega^2$, then the roots of the equation $(x-1)^3+8=0$ are:

Solution: $(x-1)^3=-8$. So $x-1$ are the cube roots of -8. Let $y = x-1$. $y^3=-8$. One root is clearly $y=-2$. The other roots are $-2\omega$ and $-2\omega^2$. So, the values for $y$ are $-2, -2\omega, -2\omega^2$. Since $x=y+1$, the roots for x are:

• $x_1 = -2+1 = -1$.
• $x_2 = -2\omega+1$.
• $x_3 = -2\omega^2+1$.

Example 17: Evaluating a Product

Question: Find the value of $(1 - \omega)(1 - \omega^2)(1 - \omega^4)(1 - \omega^5)$.

Solution: Use $\omega^3=1$, so $\omega^4=\omega$ and $\omega^5=\omega^2$. The expression is $(1-\omega)(1-\omega^2)(1-\omega)(1-\omega^2) = [(1-\omega)(1-\omega^2)]^2$. Expanding the inside: $1-\omega-\omega^2+\omega^3 = (1-(\omega+\omega^2))+1$. Since $1+\omega+\omega^2=0$, $\omega+\omega^2=-1$. So the expression is $(1-(-1))+1=3$. The final value is $3^2=9$.

Example 18: Locus with Argument

Question: If $\text{Re}(\frac{z-1}{z+1})=0$, the locus of z is:

Solution: Let $z=x+iy$. $\frac{z-1}{z+1} = \frac{(x-1)+iy}{(x+1)+iy} = \frac{((x-1)+iy)((x+1)-iy)}{((x+1)+iy)((x+1)-iy)} = \frac{(x-1)(x+1)-i(x-1)y+iy(x+1)+y^2}{(x+1)^2+y^2}$. The real part is $\frac{x^2-1+y^2}{(x+1)^2+y^2}$. Setting this to 0 gives $x^2+y^2-1=0$, which is $x^2+y^2=1$. This is a unit circle. We must exclude $z=-1$ where the expression is undefined.

Example 19: Argument of a pure imaginary number

Question: Find the principal argument of $z=-3i$.

Solution: $z=0-3i$. The point $(0, -3)$ lies on the negative imaginary axis. The angle made with the positive real axis is $-90^\circ$ or $-\pi/2$. Since this value is in the range $(-\pi, \pi]$, the principal argument is $-\pi/2$.

Example 20: Finding fourth roots of unity

Question: Find the fourth roots of unity.

Solution: We solve $z^4=1$. In polar form, $1 = \cos(0) + i\sin(0)$. The roots are $z_k = \cos(\frac{0+2k\pi}{4}) + i\sin(\frac{0+2k\pi}{4})$ for $k=0,1,2,3$.

• k=0: $z_0 = \cos(0)+i\sin(0) = 1$.
• k=1: $z_1 = \cos(\pi/2)+i\sin(\pi/2) = i$.
• k=2: $z_2 = \cos(\pi)+i\sin(\pi) = -1$.
• k=3: $z_3 = \cos(3\pi/2)+i\sin(3\pi/2) = -i$. The roots are $1, i, -1, -i$.

Example 21: Simplifying Powers of Iota

Question: Find the value of $i^{2025} + i^{-2025}$.

Solution: First, we find the remainder of 2025 when divided by 4. $2025 = 4 \times 506 + 1$. The remainder is 1. So, $i^{2025} = i^1 = i$. And $i^{-2025} = 1/i^{2025} = 1/i = -i$. The sum is $i + (-i) = 0$.

Example 22: Properties of Cube Roots of Unity

Question: If $\omega$ is a complex cube root of unity, find the value of $(1 - \omega + \omega^2)^7$.

Solution: We know $1 + \omega + \omega^2 = 0$, so $1 + \omega^2 = -\omega$. The expression becomes $(-\omega - \omega)^7 = (-2\omega)^7 = -128\omega^7$. Since $\omega^3=1$, $\omega^7 = \omega$. So the value is $-128\omega$.

Example 23: Locus of a Point (Perpendicular Bisector)

Question: Find the locus of z if $|z-2i| = |z+2|$.

Solution: This equation states that the distance of z from the point $2i$ (0,2) is equal to its distance from the point $-2$ (-2,0). The locus is the perpendicular bisector of the line segment joining these two points. The slope of the segment is $(2-0)/(0-(-2)) = 1$. The perpendicular slope is -1. The midpoint is $(-1, 1)$. The equation of the line is $y-1 = -1(x+1)$, which simplifies to $y = -x$ or $x+y=0$.

Example 24: Solving with Cube Roots of Unity

Question: If the cube roots of unity are $1, \omega, \omega^2$, then find the roots of the equation $(x-1)^3+8=0$.

Solution: $(x-1)^3=-8$. So $x-1$ are the cube roots of -8. The cube roots of -8 are $-2, -2\omega, -2\omega^2$. So, the values for $x-1$ are $-2, -2\omega, -2\omega^2$. The roots for x are: $x_1 = -2+1 = -1$, $x_2 = 1-2\omega$, $x_3 = 1-2\omega^2$.

Example 25: Product involving Cube Roots of Unity

Question: Find the value of $(1 - \omega)(1 - \omega^2)(1 - \omega^4)(1 - \omega^5)$.

Solution: Use $\omega^4=\omega$ and $\omega^5=\omega^2$. The expression is $[(1-\omega)(1-\omega^2)]^2$. Expanding the inside: $1-\omega-\omega^2+\omega^3 = (1-(\omega+\omega^2))+1$. Since $\omega+\omega^2=-1$, this is $(1-(-1))+1=3$. The final value is $3^2=9$.

Example 26: Rotation of Complex Numbers

Question: If the point $z=1+i$ is rotated counter-clockwise by an angle of $\pi/2$ about the origin, what is the new point?

Solution: Rotation by an angle $\theta$ is achieved by multiplying by $e^{i\theta}$. Here, we multiply by $e^{i\pi/2} = \cos(\pi/2)+i\sin(\pi/2) = i$. The new point $z'$ is $z' = z \cdot i = (1+i)i = i+i^2 = -1+i$.

Example 27: Sum of Roots of Unity

Question: Find the sum of the 5th roots of unity.

Solution: The n-th roots of unity are the solutions to $z^n-1=0$. By Vieta's formulas, the sum of the roots is the negative of the coefficient of the $z^{n-1}$ term, divided by the coefficient of the $z^n$ term. The equation is $z^5-1=0$. The coefficient of $z^4$ is 0. Therefore, the sum of the roots is 0.

Example 28: Maximum Value with Modulus

Question: If $|z-2|=1$, find the maximum value of $|z|$.

Solution: By the triangle inequality, $||z|-|2|| \le |z-2|$. So, $||z|-2| \le 1$. This implies $-1 \le |z|-2 \le 1$. Adding 2 to all parts gives $1 \le |z| \le 3$. The maximum value of $|z|$ is 3.

Example 29: Purely Real Condition

Question: If $\frac{z-1}{z+1}$ is a purely real number, what is the locus of z?

Solution: If a complex number $w$ is purely real, then $w = \bar{w}$. Let $w = \frac{z-1}{z+1}$. Then $\frac{z-1}{z+1} = \frac{\bar{z}-1}{\bar{z}+1}$. Cross-multiplying gives $(z-1)(\bar{z}+1) = (z+1)(\bar{z}-1)$, which simplifies to $z\bar{z}+z-\bar{z}-1 = z\bar{z}-z+\bar{z}-1$. This gives $2z=2\bar{z}$, so $z=\bar{z}$. This is true if and only if z is a real number. The locus is the real axis, excluding the point $z=-1$ where the expression is undefined.

Example 30: Solving with Complex Coefficients

Question: Find the roots of $z^2 + (1-i)z - i = 0$.

Solution: Using the quadratic formula: $D=(1-i)^2 - 4(1)(-i) = 1-2i-1+4i = 2i$. We need $\sqrt{2i}$. Let $\sqrt{2i}=x+iy$. Then $2i=x^2-y^2+2ixy$. So $x^2-y^2=0$ and $2xy=2 \implies xy=1$. Thus $x=y$. This gives $x^2=1 \implies x=\pm 1$. The roots are $1+i$ and $-1-i$. So $\sqrt{2i}=\pm(1+i)$. The roots of the equation are $z = \frac{-(1-i) \pm (1+i)}{2}$. Root 1: $\frac{-1+i+1+i}{2} = i$. Root 2: $\frac{-1+i-1-i}{2}=-1$.

Example 31: De Moivre's with Simplification

Question: Find the value of $\frac{(\cos\theta+i\sin\theta)^4}{(\sin\theta+i\cos\theta)^5}$.

Solution: The numerator is $e^{i4\theta}$. The denominator is not in standard form. We can write $\sin\theta+i\cos\theta = i(\cos\theta-i\sin\theta) = ie^{-i\theta}$. So the denominator is $(ie^{-i\theta})^5 = i^5 e^{-i5\theta} = i e^{-i5\theta}$. The expression is $\frac{e^{i4\theta}}{ie^{-i5\theta}} = \frac{1}{i} e^{i9\theta} = -i e^{i9\theta} = -i(\cos(9\theta)+i\sin(9\theta)) = \sin(9\theta)-i\cos(9\theta)$.