The correct option is D 5 = | (3+i)(2 i)1+i| =|6 3i+2i+11+i| =|7 i1+i| =|(7 i)(1 i)2| =|7 7i i 12| =|6 8i2| =|3 4i| ...

You visited us 2 times! Enjoying our articles? Unlock Full Access!

Properties of Modulus

| 3+i2-i 1+i ...

Question

$∣ ∣ ∣ \frac{(3 + i) (2 - i)}{1 + i} ∣ ∣ ∣ =$

\sqrt{5}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

5 \sqrt{2}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\sqrt{10}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

5

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is D $5$
$= | \frac{(3 + i) (2 - i)}{1 + i} |$
$= | \frac{6 - 3 i + 2 i + 1}{1 + i} |$
$= | \frac{7 - i}{1 + i} |$
$= | \frac{(7 - i) (1 - i)}{2} |$
$= | \frac{7 - 7 i - i - 1}{2} |$
$= | \frac{6 - 8 i}{2} |$
$= | 3 - 4 i |$
$= \sqrt{3^{2} + 4^{2}}$
$= \sqrt{25}$
$= 5$

Suggest Corrections

Similar questions

a > 0

z = \frac{(1 + i)^{2}}{a - i}

\sqrt{\frac{2}{5}}

¯ ¯ ¯ z

a + i b

\frac{1}{i} + \frac{2}{i^{2}} + \frac{3}{i^{3}} + \frac{5}{i^{4}}

A + i B

\frac{3 + 5 i}{2 - 3 i}

\frac{\sqrt{3} - i \sqrt{2}}{2 \sqrt{3} - i \sqrt{2}}

\frac{1 + i}{1 - i}

\frac{{(1 + i)}^{2}}{3 - i}

\frac{{(a + i b)}^{2}}{a - i b} - \frac{{(a - i b)}^{2}}{a + i b}

{[\frac{- 1 + i \sqrt{3}}{2}]}^{6} + {[\frac{- 1 - i \sqrt{3}}{2}]}^{6} + {[\frac{- 1 + i \sqrt{3}}{2}]}^{5} + {[\frac{- 1 - i \sqrt{3}}{2}]}^{5} =

{[\frac{- 1 + i \sqrt{3}}{2}]}^{6} + {[\frac{- 1 - i \sqrt{3}}{2}]}^{6} + {[\frac{- 1 + i \sqrt{3}}{2}]}^{5} + {[\frac{- 1 - i \sqrt{3}}{2}]}^{5} =

Join BYJU'S Learning Program