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for i 1 the sum of 3 4 i 4 3i is equal to
Question:
For i =
−
1
\sqrt{-1}
−
1
, the sum of
(
(
(
−
3
-3
−
3
+
+
+
4
i
i
i
)
)
)
+
+
+
(
(
(
−
4
−
3
i
)
-4-3i)
−
4
−
3
i
)
is equal to
KMAT KERALA - 2022
KMAT KERALA
Updated On:
Oct 14, 2024
−
1
−
i
-1-i
−
1
−
i
−
1
+
i
-1+i
−
1
+
i
−
7
+
i
-7+i
−
7
+
i
−
7
−
i
-7-i
−
7
−
i
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The Correct Option is
C
Solution and Explanation
To find the sum of the complex numbers
(
(
(
−
3
-3
−
3
+
+
+
4
4
4
i
i
i
)
)
)
and
(
(
(
−
4
−
3
i
)
-4-3i)
−
4
−
3
i
)
we can add the real parts and the imaginary parts separately
Real parts:
⇒
−
3
+
(
−
4
)
=
−
7
-3+(-4)=-7
−
3
+
(
−
4
)
=
−
7
Imaginary parts:
⇒
4
i
+
(
−
3
i
)
=
1
i
4i+(-3i)=1i
4
i
+
(
−
3
i
)
=
1
i
or simply
i
i
i
The sum is:
−
7
+
i
-7+i
−
7
+
i
So the correct option is (C):
−
7
+
i
-7+i
−
7
+
i
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Top Questions on complex numbers
Let
z
1
,
z
2
,
z
3
z_1, z_2, z_3
z
1
,
z
2
,
z
3
be three complex numbers on the circle
∣
z
∣
=
1
|z| = 1
∣
z
∣
=
1
with
arg
(
z
1
)
=
−
π
4
,
arg
(
z
2
)
=
0
\arg(z_1) = -\frac{\pi}{4}, \arg(z_2) = 0
ar
g
(
z
1
)
=
−
4
π
,
ar
g
(
z
2
)
=
0
and
arg
(
z
3
)
=
π
4
\arg(z_3) = \frac{\pi}{4}
ar
g
(
z
3
)
=
4
π
. If
∣
z
1
z
2
‾
+
z
2
z
3
‾
+
z
3
z
1
‾
∣
2
=
α
+
β
2
|z_1 \overline{z_2} + z_2 \overline{z_3} + z_3 \overline{z_1}|^2 = \alpha + \beta \sqrt{2}
∣
z
1
z
2
+
z
2
z
3
+
z
3
z
1
∣
2
=
α
+
β
2
, where
α
,
β
∈
Z
\alpha, \beta \in \mathbb{Z}
α
,
β
∈
Z
, then the value of
α
2
+
β
2
\alpha^2 + \beta^2
α
2
+
β
2
is :
JEE Main - 2025
Mathematics
complex numbers
View Solution
Let integers
a
,
b
∈
[
−
3
,
3
]
a, b \in [-3, 3]
a
,
b
∈
[
−
3
,
3
]
be such that
a
+
b
≠
0
a + b \neq 0
a
+
b
=
0
.
Then the number of all possible ordered pairs
\text{Then the number of all possible ordered pairs}
Then the number of all possible ordered pairs
(
a
,
b
)
(a, b)
(
a
,
b
)
,
for which
\text{for which}
for which
∣
z
−
a
z
+
b
∣
=
1
and
∣
z
+
1
ω
ω
2
ω
2
1
z
+
ω
ω
2
1
z
+
ω
∣
=
1
,
\left| \frac{z - a}{z + b} \right| = 1 \quad \text{and} \quad \left| \begin{matrix} z + 1 & \omega & \omega^2 \\ \omega^2 & 1 & z + \omega \\ \omega^2 & 1 & z + \omega \end{matrix} \right| = 1,
z
+
b
z
−
a
=
1
and
z
+
1
ω
2
ω
2
ω
1
1
ω
2
z
+
ω
z
+
ω
=
1
,
is equal to:
\text{is equal to:}
is equal to:
JEE Main - 2025
Mathematics
complex numbers
View Solution
Let the curve
z
(
1
+
i
)
+
z(1 + i) +
z
(
1
+
i
)
+
z
(
1
−
i
)
=
4
‾
\overline{z(1 - i) = 4}
z
(
1
−
i
)
=
4
, z
∈
\in
∈
C
\mathbb{C}
C
, divide the region
∣
z
−
3
∣
≤
1
|z - 3| \leq 1
∣
z
−
3∣
≤
1
into two parts of areas
α
\alpha
α
and
β
\beta
β
. Then
∣
α
−
β
∣
|\alpha - \beta|
∣
α
−
β
∣
equals:}
JEE Main - 2025
Mathematics
complex numbers
View Solution
Let O be the origin, the point A be
z
1
=
3
+
2
2
i
z_1 = \sqrt{3} + 2\sqrt{2}i
z
1
=
3
+
2
2
i
, the point B
z
2
z_2
z
2
be such that
3
∣
z
2
∣
=
∣
z
1
∣
\sqrt{3}|z_2| = |z_1|
3
∣
z
2
∣
=
∣
z
1
∣
and
arg
(
z
2
)
=
arg
(
z
1
)
+
π
6
\arg(z_2) = \arg(z_1) + \frac{\pi}{6}
ar
g
(
z
2
)
=
ar
g
(
z
1
)
+
6
π
. Then:
JEE Main - 2025
Mathematics
complex numbers
View Solution
Let the curve
(
(
(
z
(
1
+
i
)
+
z
(
1
−
i
)
‾
=
4
z(1 + i) + \overline{z(1 - i)} = 4
z
(
1
+
i
)
+
z
(
1
−
i
)
=
4
,
z
∈
C
\, z \in \mathbb{C}
z
∈
C
)
)
)
, divide the region
∣
z
−
3
∣
≤
1
|z - 3| \leq 1
∣
z
−
3∣
≤
1
into two parts of areas
α
\alpha
α
and
β
\beta
β
. Then
∣
α
−
β
∣
|\alpha - \beta|
∣
α
−
β
∣
equals:
JEE Main - 2025
Mathematics
complex numbers
View Solution
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