If - z -3-2 i -4- then sum of max - z - and min - z - is-A- 4B- 8C- NoneD- 2 -13

The correct option is B 4 = |z + ( 3+2i)| ≤ |z| + √(13) ⇒ |z| ≥ 4 √(13) Also 4 = |z+( 3+2i)| ≥ |z| √(13) ⇒ |z| ≤ 4+√(13) Hence, max nbsp;|z|+min |z| = 4 ...

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Byju's Answer

Standard XII

Mathematics

Inequality of 2 Complex Numbers

If | z -3+2 i...

Question

If $| z - 3 + 2 i |$ = 4, then sum of max|z| and min|z| is:

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Solution

The correct option is C

4 = $| z + (- 3 + 2 i) | \leq | z | + \sqrt{13}$ $\Rightarrow | z | \geq 4 - \sqrt{13}$

Also 4 = $| z + (- 3 + 2 i) | \geq | z | - \sqrt{13}$ $\Rightarrow | z | \leq 4 + \sqrt{13}$

Hence, max |z|+min |z| = $4 + \sqrt{13} + - \sqrt{13}$ = 8

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If |z−3+2i| = 4, then sum of max|z| and min|z| is:

The correct option is C 4 = |z+(−3+2i)|≤|z|+√13 ⇒|z|≥4−√13 Also 4 = |z+(−3+2i)|≥|z|−√13 ⇒|z|≤4+√13 Hence, max |z|+min |z| = 4+√13+−√13 = 8

If $| z - 3 + 2 i |$ = 4, then sum of max|z| and min|z| is:

The correct option is C

4 = $| z + (- 3 + 2 i) | \leq | z | + \sqrt{13}$ $\Rightarrow | z | \geq 4 - \sqrt{13}$

Also 4 = $| z + (- 3 + 2 i) | \geq | z | - \sqrt{13}$ $\Rightarrow | z | \leq 4 + \sqrt{13}$

Hence, max |z|+min |z| = $4 + \sqrt{13} + - \sqrt{13}$ = 8