If $a+ib=c+id$, then: (A) $a^2+c^2=0$ (B) $b^2+c^2=0$ (C) $b^2+d^2=0$ (D) $a^2+b^2=c^2+d^2$

> [!TLDR] > For two complex numbers to be equal, both their real and imaginary parts must match. This gives $a=c$ and $b=d$, which directly implies $a^2+b^2=c^2+d^2$. The correct option is **(D)**. The core idea here is simple but often rushed: equality of complex numbers is not about some vague "sameness" — it's a precise condition. A complex number $z = x + iy$ is an ordered pair $(x, y)$ of real numbers. Two such pairs are equal only when both coordinates match. That's the entire logical foundation. Let's walk through it. 1. **State the definition clearly.** If $a+ib = c+id$, where $a,b,c,d$ are real numbers, then by definition of complex equality: $$a = c \quad \text{and} \quad b = d.$$ There is no other possibility. The real parts must be equal, and the imaginary parts must be equal — separately. 2. **Check each option against this condition.** - **Option (A):** $a^2 + c^2 = 0$. Since $a = c$, this becomes $a^2 + a^2 = 2a^2 = 0$, which forces $a = 0$. But the given condition does **not** require $a = 0$ — $a$ could be any real number. So (A) is not necessarily true. - **Option (B):** $b^2 + c^2 = 0$. Here $c = a$, so this is $b^2 + a^2 = 0$, which forces $a = 0$ and $b = 0$. Again, the original condition does not force both to be zero. So (B) is false. - **Option (C):** $b^2 + d^2 = 0$. Since $b = d$, this becomes $2b^2 = 0$, forcing $b = 0$. Not required. So (C) is false. - **Option (D):** $a^2 + b^2 = c^2 + d^2$. …

If a+ib=c+id, then: (A) a²+c²=0 (B) b²+c²=0 (C) b²+d²=0 (D) a²+b²=c²+d²

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a+ib=c+id

Meghalaya MboseMCQ· 1mImportance★★★★★est

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Concept understanding — Ordered Pair Equality

Ordered Pair Equality: From Intuition to Precision

Think about a simple list of two things — say, your name and your age. If I write (Ravi, 15), that's an ordered pair. The word "ordered" is the key: the first position and the second position mean different things. (Ravi, 15) is not the same as (15, Ravi), because the first one tells you a name first, the second tells you an age first.

Now, when are two such pairs equal? Intuitively, they are equal only when both the first things match and both the second things match, in that exact order.

So (Ravi, 15) equals (Ravi, 15), but it does not equal (15, Ravi) — even though both contain the same two items. The order matters.

The Precise Statement

(a, b) = (c, d) \quad \iff \quad a = c \ \text{and} \ b = d

Read this as: "The ordered pair (a, b) equals the ordered pair (c, d) if and only if a equals c and b equals d."

Two conditions must hold simultaneously:

The first components are equal: $a = c$
The second components are equal: $b = d$

If either condition fails, the pairs are different.

Why This Matters

This definition is the foundation for everything that uses ordered pairs — coordinates in the plane, relations, functions, and even complex numbers. When you plot the point $(3, 5)$ on a graph, you are implicitly using this rule: $(3, 5)$ is a different point from $(5, 3)$ because the first coordinates differ.

Watch out

A common mistake is to think $(a, b) = (b, a)$ just because the same two objects appear. That is false unless $a = b$ . For example, $(2, 3) \neq (3, 2)$ .

Quick Check

Which of these are true?

$(4, 7) = (4, 7)$ → True (both components match)
$(4, 7) = (7, 4)$ → False (first components differ: $4 \neq 7$ )
$(x, 5) = (3, 5)$ → True only if $x = 3$
$(p, q) = (q, p)$ → True only if $p = q$

The last one surprises many students. If $p = q$ , then the pair becomes $(p, p)$ and swapping gives the same thing. But if $p \neq q$ , they are different.

One More Layer: Why "Ordered"? …

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For two complex numbers to be equal, both their real and imaginary parts must match. This gives $a=c$ and $b=d$ , which directly implies $a^2+b^2=c^2+d^2$ . The correct option is (D).

The core idea here is simple but often rushed: equality of complex numbers is not about some vague "sameness" — it's a precise condition. A complex number $z = x + iy$ is an ordered pair $(x, y)$ of real numbers. Two such pairs are equal only when both coordinates match. That's the entire logical foundation.

Let's walk through it.

State the definition clearly. If $a+ib = c+id$ , where $a,b,c,d$ are real numbers, then by definition of complex equality:

a = c \quad \text{and} \quad b = d.

There is no other possibility. The real parts must be equal, and the imaginary parts must be equal — separately.

Check each option against this condition.
- Option (A): $a^2 + c^2 = 0$ .
  
  Since $a = c$ , this becomes $a^2 + a^2 = 2a^2 = 0$ , which forces $a = 0$ . But the given condition does not require $a = 0$ — $a$ could be any real number. So (A) is not necessarily true.
- Option (B): $b^2 + c^2 = 0$ .
  
  Here $c = a$ , so this is $b^2 + a^2 = 0$ , which forces $a = 0$ and $b = 0$ . Again, the original condition does not force both to be zero. So (B) is false.
- Option (C): $b^2 + d^2 = 0$ .
  
  Since $b = d$ , this becomes $2b^2 = 0$ , forcing $b = 0$ . Not required. So (C) is false.
- Option (D): $a^2 + b^2 = c^2 + d^2$ . …

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Q.If a+ib=c+ida+ib=c+ida+ib=c+id, then: (A) a2+c2=0a^2+c^2=0a2+c2=0 (B) b2+c2=0b^2+c^2=0b2+c2=0 (C) b2+d2=0b^2+d^2=0b2+d2=0 (D) a2+b2=c2+d2a^2+b^2=c^2+d^2a2+b2=c2+d2

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Q.If $a+ib=c+id$ , then:
(A) $a^2+c^2=0$
(B) $b^2+c^2=0$
(C) $b^2+d^2=0$
(D) $a^2+b^2=c^2+d^2$

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