JEE Main 2019 (12 April)

LEVELJEE Main

Animated Solution for Mathematics - Complex Numbers: Let $z \in C$ with $I m (z) = 10$ and it satisfies $\frac{2 z - n}{2 z + n} = 2 i - 1$ for some natural number n. Then :

Select Answer:

Progressive Hints

Hint 1

The Spark: Visualize the complex number

z

as a point on the line

y = 10

in the Argand plane, so

z = x + 10 i

. The equation

\frac{2 z - n}{2 z + n} = 2 i - 1

is a relationship between

z

and

n

. Instead of trying to isolate

z

, treat this as an algebraic equation where you have two unknowns,

x

and

n

Hint 2

The Arsenal: Avoid dividing complex numbers directly, as it complicates the algebra. Instead, cross-multiply the equation to get

(2 z - n) = (2 i - 1) (2 z + n)

. This will allow you to expand both sides and equate the real and imaginary parts separately, which is the fundamental property of equality for complex numbers.

Hint 3

The Execution: Substitute

z = x + 10 i

into the cross-multiplied equation:

(2 (x + 10 i) - n) = (2 i - 1) (2 (x + 10 i) + n)

. Expand the right-hand side carefully, remembering that

i^{2} = - 1

, and then group the real and imaginary components on both sides to solve for

x

and

n

Visualized Solution

Define $z$ using $I m (z) = 10$

Given: $I m (z) = 10$
Let $z = x + 10 i$ , where $x \in R$
The point $z$ lies on the horizontal line $y = 10$ in the complex plane.

Substitute $z$ into the Equation

Given equation: $\frac{2 z - n}{2 z + n} = 2 i - 1$
Substitute $z = x + 10 i$ :
$\frac{2 ( x + 10 i ) - n}{2 ( x + 10 i ) + n} = 2 i - 1$

Simplify the Expression

Expand the terms in the numerator and denominator:
$\frac{( 2 x - n ) + 20 i}{( 2 x + n ) + 20 i} = 2 i - 1$

Cross-Multiply to Remove Fraction

Cross-multiply to clear the denominator:
$(2 x - n) + 20 i = (2 i - 1) \cdot [(2 x + n) + 20 i]$

Expand the Right-Hand Side

Distribute $(2 i - 1)$ across the bracket:
RHS $= 2 i (2 x + n) + 2 i (20 i) - 1 (2 x + n) - 1 (20 i)$

Apply $i^{2} = - 1$

Using $i^{2} = - 1$ :
RHS $= i (4 x + 2 n) + 40 (- 1) - (2 x + n) - 20 i$
RHS $= i (4 x + 2 n) - 40 - 2 x - n - 20 i$

Group Real and Imaginary Parts

Group terms on RHS:
LHS: $(2 x - n) + 20 i$
RHS: $(- 2 x - n - 40) + i (4 x + 2 n - 20)$

Equate the Real Parts

Two complex numbers are equal if their Real and Imaginary parts are equal.
Equating Real parts from LHS and RHS:
$2 x - n = - 2 x - n - 40$
Cancel $- n$ from both sides:
$2 x = - 2 x - 40$

Solve for $x$ (Real Part)

Solve for $x$ :
$4 x = - 40$
$x = - 10$
So, $R e (z) = - 10$

Equate the Imaginary Parts

Now, equate the Imaginary parts from LHS and RHS:
$20 = 4 x + 2 n - 20$
Add $20$ to both sides:
$40 = 4 x + 2 n$

Solve for $n$

Substitute $x = - 10$ :
$40 = 4 (- 10) + 2 n$
$40 = - 40 + 2 n$
$80 = 2 n$
$n = 40$

Final Conclusion

Final Values:
$n = 40$
$R e (z) = - 10$
Correct Option: n = 40 and Re(z) = -10

00:00 / 00:00

The Sigma Insight: Algebraic Operations on Complex Numbers

Solution Diagram

Analyzing the Setup

We are given a complex number

z

such that

I m (z) = 10

. This implies that

z

can be written in the form:

z = x + 10 i

where

x

is the real part we aim to determine. We are also given the governing equation involving a natural number

n

\frac{2 z - n}{2 z + n} = 2 i - 1

The Algebraic Dance

To solve for

x

and

n

, we first clear the fraction by multiplying both sides by the denominator

(2 z + n)

2 z - n = (2 i - 1) (2 z + n)

Next, we substitute

z = x + 10 i

into the equation:

2 (x + 10 i) - n = (2 i - 1) (2 (x + 10 i) + n)

Expanding the left-hand side (LHS) gives:

2 x + 20 i - n = (2 x - n) + 20 i

Expanding the Right-Hand Side

Now, we expand the right-hand side (RHS) by distributing

(2 i - 1)

R H S = (2 i - 1) ((2 x + n) + 20 i)

R H S = 2 i (2 x + n) + 2 i (20 i) - 1 (2 x + n) - 1 (20 i)

Recalling that

i^{2} = - 1

, the term

2 i (20 i)

becomes

- 40

. Thus:

R H S = (4 x i + 2 ni) - 40 - 2 x - n - 20 i

R H S = (- 2 x - n - 40) + i (4 x + 2 n - 20)

Equating Real and Imaginary Parts

By equating the real and imaginary parts of the LHS and RHS, we obtain a system of two equations. For the real parts:

2 x - n = - 2 x - n - 40

The

- n

terms cancel out, simplifying the expression to:

4 x = - 40 \Rightarrow x = - 10

For the imaginary parts, we equate the coefficients of

i

20 = 4 x + 2 n - 20

Substituting

x = - 10

into this equation:

20 = 4 (- 10) + 2 n - 20

20 = - 40 + 2 n - 20

20 = 2 n - 60

2 n = 80 \Rightarrow n = 40

Final Result

Through systematic substitution and separation of real and imaginary components, we have determined the values:

$R e (z) = - 10$ $n = 40$

Common Traps & FAQs

Wait, why can't we just divide the complex numbers directly to find z?

You absolutely could, but division of complex numbers involves multiplying by the conjugate of the denominator, which often leads to messy, error-prone expressions. Cross-multiplication is a much cleaner, more robust strategy that keeps the real and imaginary parts distinct from the start.

Is there a geometric way to solve this instead of all this algebra?

While complex numbers have beautiful geometric interpretations, this specific problem is an algebraic constraint. The algebra is actually quite elegant here because the terms involving

n

and

x

align perfectly, allowing for a straightforward solution. Sometimes, the most direct path is the algebraic one!

Why do we equate the real and imaginary parts separately?

That is the fundamental definition of equality for complex numbers! If

a + bi = c + d i

, then it must be true that

a = c

and

b = d

. This property is your most powerful tool for solving equations involving complex variables.

Animated Solution for Mathematics - Complex Numbers: Let $z \in C$ with $I m (z) = 10$ and it satisfies $\frac{2 z - n}{2 z + n} = 2 i - 1$ for some natural number n. Then :

Select Answer:

Visualized Solution

Define $z$ using $I m (z) = 10$

Substitute $z$ into the Equation

Simplify the Expression

Cross-Multiply to Remove Fraction

Expand the Right-Hand Side

Apply $i^{2} = - 1$

Group Real and Imaginary Parts

Equate the Real Parts

Solve for $x$ (Real Part)

Equate the Imaginary Parts

Solve for $n$

Final Conclusion

The Sigma Insight: Algebraic Operations on Complex Numbers

Analyzing the Setup

The Algebraic Dance

Expanding the Right-Hand Side

Equating Real and Imaginary Parts

Final Result

Similar Questions

If $z = \frac{1}{2} - 2 i$ , is such that $∣ z + 1∣ = α z + β (1 + i), i = - 1$ and $α, β \in R$ , then $α + β$ is equal to

Let $z \in C$ be such that $\frac{z ^{2} + 3 i}{z - 2 + i} = 2 + 3 i$ . Then the sum of all possible values of $z^{2}$ is

If the least and the largest real values of $α$ , for which the equation $z + α ∣ z -1∣ + 2 i = 0$ ( $z \in C$ and $i = - 1$ ) has a solution, are p and q respectively; then $4 (p^{2} + q^{2})$ is equal to

Let $z_{1}$ and $z_{2}$ be any two non-zero complex numbers such that $3∣ z_{1} ∣ = 4∣ z_{2} ∣$ . If $z = \frac{3 z _{1}}{2 z _{2}} + \frac{2 z _{2}}{3 z _{1}}$ then :

Let $S = {z \in C : z^{2} + \overset{z}{ˉ} = 0}$ . Then $\sum_{z \in S} (Re (z) + Im (z))$ is equal to

Let $z$ be a complex number such that the imaginary part of $z$ is non-zero and $a = z^{2} + z + 1$ is real. Then $a$ cannot take the value

$z \neq = 0$ is a complex number

List-I

List-II

Let $S = {z \in C : \overset{z}{ˉ} = i (z^{2} + Re (\overset{z}{ˉ}))}$ . Then $\sum_{z \in S} ∣ z ∣^{2}$ is equal to

If $(\frac{1 + i}{1 - i})^{x} = 1$ then

If $α$ and $β$ are the roots of the equation $2 z^{2} - 3 z - 2 i = 0$ where $i = - 1$ , then $16 \cdot R e (\frac{α ^{19} + β ^{19} + α ^{11} + β ^{11}}{α ^{15} + β ^{15}}) \cdot I m (\frac{α ^{19} + β ^{19} + α ^{11} + β ^{11}}{α ^{15} + β ^{15}})$ is equal to

Animated Solution for Mathematics - Complex Numbers: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let z∈C with Im(z)=10 and it satisfies 2z+n2z−n​=2i−1 for some natural number n. Then :

Select Answer:

Visualized Solution

Define .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } z using .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Im(z)=10

Substitute .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } z into the Equation

Simplify the Expression

Cross-Multiply to Remove Fraction

Expand the Right-Hand Side

Apply .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } i2=−1

Group Real and Imaginary Parts

Equate the Real Parts

Solve for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x (Real Part)

Equate the Imaginary Parts

Solve for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } n

Final Conclusion

The Sigma Insight: Algebraic Operations on Complex Numbers

Analyzing the Setup

The Algebraic Dance

Expanding the Right-Hand Side

Equating Real and Imaginary Parts

Final Result

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If z=21​−2i, is such that ∣z+1∣=αz+β(1+i),i=−1​ and α,β∈R, then α+β is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let z∈C be such that z−2+iz2+3i​=2+3i. Then the sum of all possible values of z2 is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let z1​ and z2​ be any two non-zero complex numbers such that 3∣z1​∣=4∣z2​∣. If z=2z2​3z1​​+3z1​2z2​​ then :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let S={z∈C:z2+zˉ=0}. Then ∑z∈S​(Re(z)+Im(z)) is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let z be a complex number such that the imaginary part of z is non-zero and a=z2+z+1 is real. Then a cannot take the value

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } z=0 is a complex number

List-I

List-II

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let S={z∈C:zˉ=i(z2+Re(zˉ))}. Then ∑z∈S​∣z∣2 is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If (1−i1+i​)x=1 then

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If α and β are the roots of the equation 2z2−3z−2i=0 where i=−1​, then 16⋅Re(α15+β15α19+β19+α11+β11​)⋅Im(α15+β15α19+β19+α11+β11​) is equal to

Animated Solution for Mathematics - Complex Numbers: Let $z \in C$ with $I m (z) = 10$ and it satisfies $\frac{2 z - n}{2 z + n} = 2 i - 1$ for some natural number n. Then :

Define $z$ using $I m (z) = 10$

Substitute $z$ into the Equation

Apply $i^{2} = - 1$

Solve for $x$ (Real Part)

Solve for $n$

If $z = \frac{1}{2} - 2 i$ , is such that $∣ z + 1∣ = α z + β (1 + i), i = - 1$ and $α, β \in R$ , then $α + β$ is equal to

Let $z \in C$ be such that $\frac{z ^{2} + 3 i}{z - 2 + i} = 2 + 3 i$ . Then the sum of all possible values of $z^{2}$ is

Let $z_{1}$ and $z_{2}$ be any two non-zero complex numbers such that $3∣ z_{1} ∣ = 4∣ z_{2} ∣$ . If $z = \frac{3 z _{1}}{2 z _{2}} + \frac{2 z _{2}}{3 z _{1}}$ then :

Let $S = {z \in C : z^{2} + \overset{z}{ˉ} = 0}$ . Then $\sum_{z \in S} (Re (z) + Im (z))$ is equal to

Let $z$ be a complex number such that the imaginary part of $z$ is non-zero and $a = z^{2} + z + 1$ is real. Then $a$ cannot take the value

$z \neq = 0$ is a complex number

Let $S = {z \in C : \overset{z}{ˉ} = i (z^{2} + Re (\overset{z}{ˉ}))}$ . Then $\sum_{z \in S} ∣ z ∣^{2}$ is equal to

If $(\frac{1 + i}{1 - i})^{x} = 1$ then

If $α$ and $β$ are the roots of the equation $2 z^{2} - 3 z - 2 i = 0$ where $i = - 1$ , then $16 \cdot R e (\frac{α ^{19} + β ^{19} + α ^{11} + β ^{11}}{α ^{15} + β ^{15}}) \cdot I m (\frac{α ^{19} + β ^{19} + α ^{11} + β ^{11}}{α ^{15} + β ^{15}})$ is equal to