If $z_1$ and $z_2$ are complex numbers such that $z_1+z_2$ is a real number, then $z_2=$ _____.

If z₁ and z₂ are complex numbers such that z₁+z₂ is a real number, then z₂= ___.

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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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If $z_1+z_2$ is real, only the imaginary parts must cancel — the real part of $z_2$ is completely unrestricted. So $z_2=\overline{z_1}+k$ for any real number $k$ ; it is not forced to equal $\overline{z_1}$ exactly.

Why this isn't just "the conjugate"

A common reflex is to say: if $z_1+z_2$ is real, then $z_2$ must be the conjugate of $z_1$ . That's not always true. Let's see why.

Write $z_1=a+bi$ and $z_2=c+di$ , where $a,b,c,d$ are real numbers. Their sum is:

z_1+z_2 = (a+c)+(b+d)i

For this to be real, the imaginary part must be zero:

b+d=0 \implies d=-b

That's the only condition. The real part $a+c$ can be any real number — there's no restriction linking $c$ to $a$ .

So $z_2=c-bi$ , where $c$ is any real number. That's the full family of possibilities.

Watch out

Don't assume $z_2=\overline{z_1}$ unless you also know $z_1$ itself is real. If $z_1$ is real, then $b=0$ , so $d=0$ too, and $z_2$ must be real — but that's a special case, not the general one.

Step-by-step reasoning

Set up the general form. Let $z_1=a+bi$ , $z_2=c+di$ , with $a,b,c,d\in\mathbb{R}$ .
Write the sum.

z_1+z_2 = (a+c)+(b+d)i

Apply the realness condition. A complex number is real if and only if its imaginary part is $0$ :

b+d=0 \implies d=-b

Interpret the result. The real part $c$ is free — it can be any real number. So: …

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Q.If z1z_1z1​ and z2z_2z2​ are complex numbers such that z1+z2z_1+z_2z1​+z2​ is a real number, then z2=z_2=z2​= _____.

Why this isn't just "the conjugate"

Step-by-step reasoning

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Q.If $z_1$ and $z_2$ are complex numbers such that $z_1+z_2$ is a real number, then $z_2=$ _____.

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