Difference between revisions of "1988 AHSME Problems/Problem 21"
(Created page with "==Problem== The complex number <math>z</math> satisfies <math>z + |z| = 2 + 8i</math>. What is <math>|z|^{2}</math>? Note: if <math>z = a + bi</math>, then <math>|z| = \sqrt{a^{...") |
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==Solution== | ==Solution== | ||
− | + | Let the complex number <math>z</math> equal <math>a+bi</math>. Then the preceding equation can be expressed as <cmath>a+bi+\sqrt{a^2+b^2} = 2+8i</cmath> Because <math>a</math> and <math>b</math> must both be real numbers, we immediately have that <math>bi = 8i</math>, giving <math>b = 8</math>. Plugging this in back to our equation gives us <math>a+\sqrt{a^2+64} = 2</math>. | |
+ | Rearranging this into <math>2-a = \sqrt{a^2+64}</math>, we can square each side of the equation resulting in <cmath>4-4a+a^2 = a^2+64</cmath> Further simplification will yield | ||
+ | <math>60 = -4a</math> meaning that <math>-15 = a</math>. Knowing both <math>a</math> and <math>b</math>, we can plug them in into <math>a^2+b^2</math>. Our final answer is <math>\boxed{289}</math>. | ||
== See also == | == See also == |
Latest revision as of 20:58, 8 June 2016
Problem
The complex number satisfies . What is ? Note: if , then .
Solution
Let the complex number equal . Then the preceding equation can be expressed as Because and must both be real numbers, we immediately have that , giving . Plugging this in back to our equation gives us . Rearranging this into , we can square each side of the equation resulting in Further simplification will yield meaning that . Knowing both and , we can plug them in into . Our final answer is .
See also
1988 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 20 |
Followed by Problem 22 | |
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All AHSME Problems and Solutions |
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