Show that the area of the triangle on the Argand diagram formed by the complex numbers z, iz and z +

Q: Wait, why is the angle between $z$ and $iz$ exactly $90^{\circ}$?

Great question! Multiplying any complex number by $i$ is equivalent to multiplying by $e^{i\pi/2}$. In polar form, this adds $\pi/2$ (or $90^{\circ}$) to the argument of the complex number without changing its magnitude. Thus, $iz$ is always a $90^{\circ}$ counter-clockwise rotation of $z$.

Q: Can I use the coordinate geometry formula for the area of a triangle with vertices $(x_1, y_1), (x_2, y_2), (x_3, y_3)$?

Absolutely! You could represent $z$ as $x+iy$, then $iz$ becomes $-y+ix$, and $z+iz$ becomes $(x-y) + i(x+y)$. Plugging these into the determinant formula for area will yield the same result. However, the geometric approach is much faster and builds better intuition for complex numbers.

Q: Why is the shape formed by $0, z, iz,$ and $z+iz$ a square and not just a rectangle?

Because the magnitude of $z$ is equal to the magnitude of $iz$ (since $|i|=1$). A parallelogram with equal adjacent sides is a rhombus, and since the angle between them is $90^{\circ}$, it must be a square.

Q: Does the area change if $z$ is in a different quadrant?

Not at all! The magnitude $|z|$ is always a positive real number, and the rotation property of $i$ is universal. The geometry remains invariant regardless of where $z$ sits in the complex plane.

Q: Is it possible for the area to be zero?

Yes, if $z=0$. If $z=0$, then $iz=0$ and $z+iz=0$. All three vertices collapse to the origin, forming a degenerate triangle with an area of zero, which is consistent with our formula $\frac{1}{2}|z|^2$.

Question

Show that the area of the triangle on the Argand diagram formed by the complex numbers $z, iz$ and $z + iz$ is $\frac{1}{2}|z|^2$.

Sigma Percentile · Accepted Answer

### The Elegant Geometry of the Complex Plane

Welcome, fellow traveler of the mathematical landscape. Today, we are going to peel back the curtain on one of the most beautiful intersections of algebra and geometry: the Argand plane.

Often, students view complex numbers as mere algebraic entities—things to be added, subtracted, or multiplied. But today, we see them as vectors, as points in space, and as participants in a geometric dance.

### The Dance of the Imaginary Unit

Imagine you have a complex number $z$ sitting somewhere in the Argand plane. It is a vector reaching out from the origin $O(0,0)$ with a magnitude of $|z|$.

Now, what happens when we multiply this vector by $i$? In the realm of complex numbers, multiplying by $i$ is not just a scaling operation; it is a rotation. Specifically, it is a $90^{\circ}$ counter-clockwise rotation.

Because the magnitude of $i$ is exactly $1$, the length of our vector remains unchanged: $|iz| = |i| \cdot |z| = |z|$. We have effectively created a new point, $iz$, which is perpendicular to $z$ and shares the same distance from the origin.

### Constructing the Square

Now, consider the sum $z + iz$. If you recall your vector addition, this is the diagonal of a parallelogram formed by $z$ and $iz$.

But wait—because the angle between $z$ and $iz$ is $90^{\circ}$ and their magnitudes are equal, this is no ordinary parallelogram. It is a perfect square! The vertices of this square are $0, z, z+iz,$ and $iz$.

This realization is the 'Aha!' moment that simplifies everything. We are not dealing with a random triangle; we are dealing with a specific slice of a square.

### Isolating the Triangle

The problem asks us to find the area of the triangle formed by the vertices $z, iz,$ and $z+iz$. Let us look at the sides of this triangle.

The vector connecting $z$ to $z+iz$ is simply $(z+iz) - z = iz$. The magnitude of this side is $|iz|$, which we already know is $|z|$.

Similarly, the vector connecting $iz$ to $z+iz$ is $(z+iz) - iz = z$. The magnitude of this side is $|z|$.

### The Final Synthesis

We have discovered that our triangle has two sides of length $|z|$ that meet at a $90^{\circ}$ angle. This is a right-angled triangle!

The area of any right-angled triangle is given by the classic formula:

$$	ext{Area} = \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height}$$

Substituting our side lengths, we get:

$$	ext{Area} = \frac{1}{2} 	imes |z| 	imes |iz|$$

Since $|iz| = |z|$, the expression simplifies beautifully to:

$$	ext{Area} = \frac{1}{2} 	imes |z| 	imes |z| = \frac{1}{2}|z|^2$$

Isn't it marvelous? What started as a seemingly abstract algebraic expression transformed into a clear, geometric truth.

You have successfully navigated the Argand plane, identified the hidden square, and calculated the area with precision. The final result is **$\frac{1}{2}|z|^2$**.

Keep this geometric intuition close; it will serve you well in the most challenging problems of the JEE Advanced. You are not just solving equations; you are uncovering the architecture of the universe.

Animated Solution for Mathematics - Complex Numbers: Show that the area of the triangle on the Argand diagram formed by the complex numbers $z, i z$ and $z + i z$ is $\frac{1}{2} ∣ z ∣^{2}$ .

Visualized Solution

Visualizing $z$ on the Argand Plane

The Effect of Multiplying by $i$

Magnitude of $i z$

Finding the Third Vertex $z + i z$

The Resultant Vector

Identifying the Target Triangle

Analyzing the Triangle's Sides

Length of the First Side

Analyzing the Second Side

Length of the Second Side

The Angle Between the Sides

Area of the Right-Angled Triangle

Final Calculation

The Sigma Insight: Geometrical Applications of Complex Numbers

The Elegant Geometry of the Complex Plane

The Dance of the Imaginary Unit

Constructing the Square

Isolating the Triangle

The Final Synthesis

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Animated Solution for Mathematics - Complex Numbers: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Show that the area of the triangle on the Argand diagram formed by the complex numbers z,iz and z+iz is 21​∣z∣2.