If Re-a- space Re-b- - 0- and x - -a - b- - -bar-a- - b- thenx0

Question

Toppr Admin · Accepted Answer

Lets take a-a1-ia2-b-b1-ib2x-a1-ia2-x2212-b1-ib2-a1-x2212-ia2-b1-ib2-So we can see from above equation-x2223-x2223-xAF-xAF-xAF-a-b-x2223-x2223-gt-a-x2212-b-Hence x-lt-0

If $R e (a), R e (b) > 0$ , and $x = | a - b | - | ¯ a + b |$ , then
$x < 0$
$x \geq 1$
0
$x > 0$

Lets take $a = a_{1} + i a_{2}, b = b_{1} + i b_{2}$
$x = | (a_{1} + i a_{2}) - (b_{1} + i b_{2}) | + | (a_{1} - i a_{2}) + (b_{1} + i b_{2}) |$
So we can see from above equation
$∣ ∣ ¯ ¯ ¯ a + b ∣ ∣ > | a - b |$
Hence $x < 0$

If Re(a), Re(b)>0, and x=|a−b|−|¯a+b|, thenx<0x≥10x>0

Lets take a=a1+ia2,b=b1+ib2x=|(a1+ia2)−(b1+ib2)|+|(a1−ia2)+(b1+ib2)|So we can see from above equation∣∣¯¯¯a+b∣∣>|a−b|Hence x<0

If $R e (a), R e (b) > 0$ , and $x = | a - b | - | ¯ a + b |$ , then
$x < 0$
$x \geq 1$
0
$x > 0$

Lets take $a = a_{1} + i a_{2}, b = b_{1} + i b_{2}$
$x = | (a_{1} + i a_{2}) - (b_{1} + i b_{2}) | + | (a_{1} - i a_{2}) + (b_{1} + i b_{2}) |$
So we can see from above equation
$∣ ∣ ¯ ¯ ¯ a + b ∣ ∣ > | a - b |$
Hence $x < 0$