-z1-z2-z1-z2- is possible ifa z 2- z - 1b z 2-1 z 1c z1-z2d -z1-z2-

Since given |z1 + z2| = |z1| + |z2| i.e |z1 + z2|2 = (|z1| + |z2|)2 i.e #160;z1+z22=z1+z22Since, #160;z1=r1, #160;z2=r2 #160; #160;where #160;z1=r1 ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard X

Mathematics

Pythagoras Theorem

|z1+z2|=|z1|+...

Question

|z₁ + z₂| = |z₁| + |z₂| is possible if

(a) $z_{2} = {\bar{z}}_{1}$

(b) $z_{2} = \frac{1}{z_{1}}$

(c) arg (z₁) = arg (z₂)

(d) |z₁| = |z₂|

Open in App

Solution

Since given |z₁ + z₂| = |z₁| + |z₂|
i.e |z₁ + z₂|² = (|z₁| + |z₂|)²
$i . e {|z_{1} + z_{2}|}^{2} = {(|z_{1}| + |z_{2}|)}^{2} Since, |z_{1}| = r_{1}, |z_{2}| = r_{2} where z_{1} = r_{1} {e^{i θ}}_{1} z_{2} = r_{2} {e^{i θ}}_{2} {|z_{1}|}^{2} + {|z_{2}|}^{2} + 2 |z_{1}| |z_{2}| \cos (θ_{1} - θ_{2}) = r_{1}^{2} + r_{2}^{2} + 2 r_{1} r_{2} i . e r_{1}^{2} + r_{2}^{2} + 2 r_{1} r_{2} \cos (θ_{1} - θ_{2}) = r_{1}^{2} + r_{2}^{2} + 2 r_{1} r_{2} i . e 2 r_{1} r_{2} \cos (θ_{1} - θ_{2}) = 2 r_{2} i . e \cos (θ_{1} - θ_{2}) = 1 i . e θ_{1} - θ_{2} = 0 i . e c h g z_{1} = c h g z_{2}$
Hence, the correct answer is option C.

Suggest Corrections

Similar questions

Q. If for a complex number

z_{1}

and

z_{2}

arg (z_{1}) - arg (z_{2}) = 0

, then

| z_{1} - z_{2} |

is equal to

| z - z_{1} |^{2} + | z - z_{2} |^{2} = a

will represent a real circle on the argand plane if

| z_{1} + z_{2} | = | z_{1} | + | z_{2} |

is possible if

Q. Which of the following is correct for any two complex numbers z₁ and z₂?

(a)

|z_{1} z_{2}| = |z_{1}| |z_{2}|

(b)

\arg (z_{1} z_{2}) = \arg (z_{1}) \arg (z_{2})

(c)

|z_{1} + z_{2}| = |z_{1}| + |z_{2}|

(d)

|z_{1} + z_{2}| \geq |z_{1}| + |z_{2}|

Q. If

z_{1}, z_{2}, z_{3}

represent the vertices of an equilateral triangle such that

| z_{1} | = | z_{2} | = | z_{3} |

, then

Join BYJU'S Learning Program

|z1 + z2| = |z1| + |z2| is possible if (a) z2=z¯1 (b) z2=1z1 (c) arg (z1) = arg (z2) (d) |z1| = |z2|

|z₁ + z₂| = |z₁| + |z₂| is possible if

(a) $z_{2} = {\bar{z}}_{1}$

(b) $z_{2} = \frac{1}{z_{1}}$

(c) arg (z₁) = arg (z₂)

(d) |z₁| = |z₂|