If Re(1+iy)3=-26, where y is a real number ,then the value of |y| is

Question

If Re(1+iy)3=-26, where y is a real number ,then the value of |y| is (A) 2 (B) 3 (C) 4 (D) 6

Tardigrade Admin · Accepted Answer

Let z=(1+i y)3 =1+3 i y+3 i2 y2+i3 y3 =1+3 i y-3 y2-i y3 (1-3 y2)+(3 y-y3) i Now, operatornameRe(z)=-26 ⇒ 1-3 y2 =-26 ⇒ 27=3 y2 ⇒ y2=9 ⇒ y=± 3 or |y|=3

Q. If $R e (1 + i y)^{3} = - 26,$ where $y$ is a real number ,then the value of $∣ y ∣$ is

2

3

4

6

9

Let $z = (1 + i y)^{3}$
$= 1 + 3 i y + 3 i^{2} y^{2} + i^{3} y^{3}$
$= 1 + 3 i y - 3 y^{2} - i y^{3}$
$(1 - 3 y^{2}) + (3 y - y^{3}) i$
Now, $Re (z) = - 26$
$\Rightarrow 1 - 3 y^{2} = - 26$
$\Rightarrow 27 = 3 y^{2} \Rightarrow y^{2} = 9$
$\Rightarrow y = \pm 3$ or $∣ y ∣ = 3$

Q. If Re(1+iy)3=−26, where y is a real number ,then the value of ∣y∣ is

2

3

4

6

9

Let z=(1+iy)3 =1+3iy+3i2y2+i3y3 =1+3iy−3y2−iy3 (1−3y2)+(3y−y3)i Now, Re(z)=−26 ⇒1−3y2=−26 ⇒27=3y2⇒y2=9 ⇒y=±3 or ∣y∣=3

Q. If $R e (1 + i y)^{3} = - 26,$ where $y$ is a real number ,then the value of $∣ y ∣$ is

Let $z = (1 + i y)^{3}$
$= 1 + 3 i y + 3 i^{2} y^{2} + i^{3} y^{3}$
$= 1 + 3 i y - 3 y^{2} - i y^{3}$
$(1 - 3 y^{2}) + (3 y - y^{3}) i$
Now, $Re (z) = - 26$
$\Rightarrow 1 - 3 y^{2} = - 26$
$\Rightarrow 27 = 3 y^{2} \Rightarrow y^{2} = 9$
$\Rightarrow y = \pm 3$ or $∣ y ∣ = 3$