mod(Re(z)) mod(Im(z))=4

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If z be0:06:46

If z be a complex number satisfying |Re(z)| + |Im(z)| =4, then |z| can’t be #IIT JEE complex numbers

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If z be a complex number satisfying |Re(z)|+|lm(z)|=4, then |z| cannot be : [JEE Main-2020 (Janua...

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Find Locus of0:10:27

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Exercice : Equation Re(z**3)=Im(z**3)

5.The inequality |z-4|0:02:43

5.The inequality |z-4| less |z-2| represents a)Re(z) 0 b) Re(z) 0 c) Re(z) 2 d) Re(z) 3

Prove that Re(iz)0:02:29

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If `|z-2|=2|z-1|`, then0:02:29

If `|z-2|=2|z-1|`, then show that `|z|^(2)=(4 )/(3)Re(z)`.

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Given z=(1+i √(3))^100,0:05:12

Given z=(1+i √(3))^100, then [RE(z) / IM(z)] equals (1) 2^100 (2) 2^50 (3) 1/√(3) (4) √(3)

Let `A={z:'Im'(z) ge0:05:44

Let `A={z:'Im'(z) ge 1}, B={z:|z-2-i|=3}, C={z:'Re'{(1-i)z}=sqrt(2)}` be three sides of

Find Locus of0:09:32

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The number of0:03:37

The number of solutions of the system of equations Re(z^2)=0, |z|=2 is

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Let z be0:12:16

Let z be a complex number such that |z+2|=1 and Im(z+1/z+2)=1/5. Then the value of |Re(z+2)| is ?

`|z-i|lt|z+i|` represents the0:01:58

`|z-i|lt|z+i|` represents the region (A) `Re(z)gt0` (B) `Re(z)lt0` (C) `Im(z)gt0` (D) `Im(z)lt0`

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If z be0:09:51

If z be a complex number satisfying |Re(z)|+|Im(z)|=4, then |z| cannot be (1) √(17/2) (2) √(10) (...

If z is0:13:02

If z is a complex numbers such that mod(z-1)=mod(z+1) show that Re(z)=0

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