If $\left(\dfrac{1-i}{1+i}\right)^{100}=a+ib$, then find $(a, b)$.

> [!TLDR] > The key idea is to simplify the complex fraction $\frac{1-i}{1+i}$ to its pure imaginary form $-i$, then raise it to the 100th power. Since $(-i)^{100} = 1$, we get $(a, b) = (1, 0)$. ## Why This Approach Works When you see a complex number raised to a high power, the worst thing you can do is multiply it out 100 times. The smart move is to simplify the base first. Here, $\frac{1-i}{1+i}$ looks like a ratio of two complex numbers — and the standard trick is to rationalise the denominator by multiplying numerator and denominator by the conjugate of the denominator. That turns the fraction into something much simpler, often a pure real or pure imaginary number. Once you have that, the power becomes trivial. Let’s walk through it. --- 1. **Rationalise the fraction** Multiply numerator and denominator by the conjugate of $1+i$, which is $1-i$: \[ \frac{1-i}{1+i} \times \frac{1-i}{1-i} = \frac{(1-i)^2}{(1+i)(1-i)} \] 2. **Simplify the denominator** $(1+i)(1-i) = 1^2 - i^2 = 1 - (-1) = 2$. So the denominator becomes $2$. 3. **Simplify the numerator** $(1-i)^2 = 1 - 2i + i^2 = 1 - 2i - 1 = -2i$. 4. **Put it together** \[ \frac{-2i}{2} = -i \] So $\frac{1-i}{1+i} = -i$. That’s a pure imaginary number — specifically, the point $(0, -1)$ on the complex plane. > [!TIP] > Once you see the base is $-i$, you’re done with heavy algebra. The rest is just pattern recognition: powers of $i$ cycle every 4, and the negative sign just flips the sign. 5. **Raise to the 100th power** We need $(-i)^{100}$. Write this as $[(-1) \cdot i]^{100} = (-1)^{100} \cdot i^{100}$. 6. **Handle the powers separately** …

If ((1-i)/(1+i))¹⁰⁰=a+ib, then find (a, b).

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\left(\dfrac{1-i}{1+i}\right)^{100}=a+ib

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Concept understanding — Complex Number Arithmetic

Complex Number Arithmetic: A First Look

Imagine you're trying to solve $x^2 + 1 = 0$ . You know that no real number squared gives $-1$ . The square of any real number is either zero or positive. So this equation has no real solution. But what if we invent a number whose square is $-1$ ? That's exactly what mathematicians did — and that invention is the imaginary unit $i$ , defined by:

i^2 = -1

A complex number is any number of the form $a + bi$ , where $a$ and $b$ are real numbers. Here $a$ is called the real part, and $b$ is called the imaginary part. For example, $3 + 4i$ has real part $3$ and imaginary part $4$ .

Note

The name "imaginary" is unfortunate — these numbers are just as real (in the mathematical sense) as the numbers you already know. They're simply a different kind of number.

Why Bother?

Complex numbers let you solve equations that real numbers can't. Every polynomial equation — no matter how complicated — has a solution in the complex numbers. This is the Fundamental Theorem of Algebra, and it's one of the most important results in mathematics.

Arithmetic Operations

The rules are straightforward: treat $i$ like a variable, but remember that $i^2 = -1$ .

Addition and Subtraction

Add (or subtract) real parts with real parts, imaginary parts with imaginary parts.

(a + bi) + (c + di) = (a + c) + (b + d)i

(a + bi) - (c + di) = (a - c) + (b - d)i

Example: $(2 + 3i) + (4 - 5i) = (2 + 4) + (3 - 5)i = 6 - 2i$

Multiplication

Multiply like binomials, then replace $i^2$ with $-1$ .

(a + bi)(c + di) = ac + adi + bci + bdi^2 = (ac - bd) + (ad + bc)i

Example: $(2 + 3i)(4 - 5i) = 2(4) + 2(-5i) + 3i(4) + 3i(-5i)$

$= 8 - 10i + 12i - 15i^2$

$= 8 + 2i - 15(-1)$

$= 8 + 2i + 15 = 23 + 2i$

Watch out

The most common mistake: forgetting that $i^2 = -1$ , not $1$ . Always check your final step.

Division

Division is trickier. The key idea: multiply numerator and denominator by the complex conjugate of the denominator.

The complex conjugate of $a + bi$ is $a - bi$ . When you multiply a complex number by its conjugate, you get a real number:

(a + bi)(a - bi) = a^2 - (bi)^2 = a^2 - b^2 i^2 = a^2 + b^2

So to divide:

\frac{a + bi}{c + di} = \frac{(a + bi)(c - di)}{(c + di)(c - di)} = \frac{(a + bi)(c - di)}{c^2 + d^2}

Example: $\frac{2 + 3i}{4 - 5i} = \frac{(2 + 3i)(4 + 5i)}{(4 - 5i)(4 + 5i)} = \frac{8 + 10i + 12i + 15i^2}{16 + 25} = \frac{8 + 22i - 15}{41} = \frac{-7 + 22i}{41} = -\frac{7}{41} + \frac{22}{41}i$

Tip

To divide quickly: multiply top and bottom by the conjugate of the denominator, then simplify. The denominator always becomes $c^2 + d^2$ , a positive real number.

The Big Picture …

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The key idea is to simplify the complex fraction $\frac{1-i}{1+i}$ to its pure imaginary form $-i$ , then raise it to the 100th power. Since $(-i)^{100} = 1$ , we get $(a, b) = (1, 0)$ .

Why This Approach Works

When you see a complex number raised to a high power, the worst thing you can do is multiply it out 100 times. The smart move is to simplify the base first. Here, $\frac{1-i}{1+i}$ looks like a ratio of two complex numbers — and the standard trick is to rationalise the denominator by multiplying numerator and denominator by the conjugate of the denominator. That turns the fraction into something much simpler, often a pure real or pure imaginary number. Once you have that, the power becomes trivial.

Let’s walk through it.

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

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

Simplify the denominator

$(1+i)(1-i) = 1^2 - i^2 = 1 - (-1) = 2$ .

So the denominator becomes $2$ .
Simplify the numerator

$(1-i)^2 = 1 - 2i + i^2 = 1 - 2i - 1 = -2i$ .
Put it together

\frac{-2i}{2} = -i

So $\frac{1-i}{1+i} = -i$ . That’s a pure imaginary number — specifically, the point $(0, -1)$ on the complex plane.

Tip

Once you see the base is $-i$ , you’re done with heavy algebra. The rest is just pattern recognition: powers of $i$ cycle every 4, and the negative sign just flips the sign.

Raise to the 100th power

We need $(-i)^{100}$ . Write this as $[(-1) \cdot i]^{100} = (-1)^{100} \cdot i^{100}$ .
Handle the powers separately …

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Q.If (1−i1+i)100=a+ib\left(\dfrac{1-i}{1+i}\right)^{100}=a+ib(1+i1−i​)100=a+ib, then find (a,b)(a, b)(a,b).

Why This Approach Works

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Q.If $\left(\dfrac{1-i}{1+i}\right)^{100}=a+ib$ , then find $(a, b)$ .

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